Health Education Systems Inc (HESI) A2 Exam Quiz: Work Energy And Power
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Work Energy And PowerQuestion 1 of 20

A nurse slowly lowers a 12 kg box of medical supplies from a 1.5 m high shelf to the floor. How much work is done by the nurse on the box?

-180 J
-18 J
18 J
180 J
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Work Energy And Power

Practice Work Energy And Power in Health Education Systems Inc (HESI) A2 Exam with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Work Energy And Power, giving you a quick way to practice the rules, question types, and explanations that matter most for Health Education Systems Inc (HESI) A2 Exam.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A nurse slowly lowers a 12 kg box of medical supplies from a 1.5 m high shelf to the floor. How much work is done by the nurse on the box?

  1. -180 J (correct answer)
  2. -18 J
  3. 18 J
  4. 180 J
Explanation: When you encounter physics problems involving work and energy in healthcare settings, remember that work depends on both the force applied and the direction of movement relative to that force. To find the work done by the nurse, you need to identify the force she applies and the displacement. As the nurse "slowly lowers" the box, she's controlling its descent against gravity. The box weighs mg=12 kg×9.8 m/s2=117.6 Nmg = 12 \text{ kg} \times 9.8 \text{ m/s}^2 = 117.6 \text{ N} downward. To lower it slowly (at constant velocity), the nurse must apply an upward force of 117.6 N. The displacement is 1.5 m downward, while the nurse's force is upward. Since force and displacement are in opposite directions, the work is negative: W=F×d×cos⁡(180°)=117.6×1.5×(−1)=−176.4 JW = F \times d \times \cos(180°) = 117.6 \times 1.5 \times (-1) = -176.4 \text{ J}. Rounding to significant figures gives approximately -180 J, confirming answer A is correct. Answer B (-18 J) likely results from using an incorrect gravitational constant or mathematical error. Answer C (18 J) represents the same magnitude error as B but with the wrong sign—missing that the nurse's force opposes the displacement. Answer D (180 J) has the correct magnitude but wrong sign, indicating a misunderstanding of the force direction relative to displacement. Remember: work is negative when the applied force opposes the direction of motion. In healthcare, you'll often encounter controlled movements where workers apply forces against gravity, resulting in negative work done by the person.

Question 2

A 1 kg textbook is dropped from a 1.25 m high counter. Assuming no air resistance, what is its approximate speed just before it hits the floor? (Use g ≈ 10 m/s²)

  1. 1.25 m/s
  2. 2.5 m/s
  3. 5.0 m/s (correct answer)
  4. 12.5 m/s
Explanation: When you encounter physics problems involving objects falling under gravity, you're dealing with kinematics and energy conservation. The key insight is that gravitational potential energy converts to kinetic energy as the object falls. You can solve this using the kinematic equation: v2=v02+2ghv^2 = v_0^2 + 2gh, where vv is final velocity, v0v_0 is initial velocity (0 for a dropped object), gg is gravitational acceleration, and hh is height. Substituting the values: v2=02+2(10)(1.25)=25v^2 = 0^2 + 2(10)(1.25) = 25, so v=5.0v = 5.0 m/s. Alternatively, use energy conservation: potential energy mghmgh equals kinetic energy 12mv2\frac{1}{2}mv^2. The mass cancels out: gh=12v2gh = \frac{1}{2}v^2, giving v=2gh=2(10)(1.25)=5.0v = \sqrt{2gh} = \sqrt{2(10)(1.25)} = 5.0 m/s. Choice A (1.25 m/s) represents a common error of confusing height with velocity or using an incorrect formula. Choice B (2.5 m/s) might result from forgetting the factor of 2 in the kinematic equation or taking the square root incorrectly. Choice D (12.5 m/s) could come from multiplying g×hg \times h directly without proper kinematic analysis. Notice that the 1 kg mass is irrelevant—gravitational acceleration affects all objects equally regardless of mass. For HESI physics questions, always identify what information is given versus what's actually needed, and remember that falling object problems depend only on height and gravity, not mass.

Question 3

A hospital porter carries a 15 kg medical supply bag at a constant velocity for a horizontal distance of 10 meters across a level floor. How much work is done on the bag by the porter during this process?

  1. 0 J (correct answer)
  2. 150 J
  3. 1,470 J
  4. 1,500 J
Explanation: When you encounter physics problems involving work, remember that work is defined as force applied in the direction of displacement: W=F⋅d⋅cos⁡(θ)W = F \cdot d \cdot \cos(\theta), where θ is the angle between force and displacement. In this scenario, the porter carries the bag at constant velocity across a horizontal surface. This is the key insight: at constant velocity, there's no net acceleration, meaning no net force acts on the bag. The porter applies an upward force to support the bag's weight (15 kg × 9.8 m/s² = 147 N), but this force is perpendicular to the horizontal displacement. Since the angle between the supporting force and horizontal motion is 90°, and cos(90°) = 0, the work done is: W=147 N×10 m×cos⁡(90°)=147×10×0=0 JW = 147 \text{ N} \times 10 \text{ m} \times \cos(90°) = 147 \times 10 \times 0 = 0 \text{ J} Answer A (0 J) is correct because no force acts in the direction of motion. Answer B (150 J) incorrectly assumes a horizontal force of 15 N over 10 meters, confusing mass with force. Answer C (1,470 J) mistakenly multiplies the bag's weight (147 N) by the distance, ignoring that weight acts vertically while motion is horizontal. Answer D (1,500 J) appears to multiply mass (15 kg) by distance and gravity, showing confusion about the work formula. Remember: work only occurs when force has a component in the direction of motion. Carrying objects horizontally at constant speed requires no horizontal force, so no work is done on the object.

Question 4

Nurse A lifts a 20 kg patient transfer board 1.5 meters from the floor to a bed in 2 seconds. Nurse B lifts an identical board the same vertical distance in 3 seconds. Which statement correctly compares the work done and power exerted?

  1. Nurse A does more work but exerts less power than Nurse B.
  2. Nurse B does more work and exerts more power than Nurse A.
  3. Both nurses do the same amount of work, but Nurse A exerts more power. (correct answer)
  4. Both nurses do the same amount of work, but Nurse B exerts more power.
Explanation: When you encounter physics problems in healthcare contexts, you need to distinguish between work and power - two related but distinct concepts. Work depends only on force and distance, while power also considers time. To find the work done, use the formula: Work = Force × Distance. Both nurses lift the same 20 kg board the same vertical distance of 1.5 meters against gravity. The force required is the weight of the board (mass × gravity), which is identical for both nurses. Since force and distance are the same in both cases, both nurses perform exactly the same amount of work. Power, however, is work divided by time: Power = Work ÷ Time. Since both nurses do the same work but Nurse A completes it in 2 seconds while Nurse B takes 3 seconds, Nurse A exerts more power. Think of it this way: doing the same job faster requires more power. Option A is incorrect because both nurses do equal work, not different amounts. Option B is wrong on both counts - the work is equal, and Nurse B actually exerts less power due to the longer time. Option D incorrectly states that Nurse B exerts more power when the opposite is true. For HESI physics questions, remember this key distinction: work depends on what you accomplish (force × distance), while power depends on how quickly you accomplish it (work ÷ time). Time affects power but never affects the total work done.

Question 5

A physical therapist applies a force of 150 N to a patient's leg, moving it 0.4 meters. The force is applied at an angle of 30° to the direction of motion. How much work is done by the therapist on the patient's leg? (cos(30°) ≈ 0.87)

  1. 30 J
  2. 52 J (correct answer)
  3. 60 J
  4. 75 J
Explanation: When you encounter physics problems involving forces and motion in healthcare contexts, you're dealing with work calculations. Work depends not just on the force applied and distance moved, but critically on the angle between the force direction and motion direction. To find work done, use the formula: W=F×d×cos⁡(θ)W = F \times d \times \cos(\theta), where F is force, d is displacement, and θ is the angle between force and motion direction. Here, the therapist applies 150 N of force over 0.4 meters at a 30° angle to the motion direction. Substituting: W=150 N×0.4 m×cos⁡(30°)=150×0.4×0.87=52.2 JW = 150 \text{ N} \times 0.4 \text{ m} \times \cos(30°) = 150 \times 0.4 \times 0.87 = 52.2 \text{ J}. This rounds to 52 J, making B correct. Choice A (30 J) likely comes from incorrectly using sin(30°) ≈ 0.5 instead of cos(30°), giving 150 × 0.4 × 0.5 = 30 J. Choice C (60 J) represents the common mistake of ignoring the angle entirely and calculating 150 × 0.4 = 60 J. Choice D (75 J) might result from calculation errors or misapplying the angle correction. The key trap here is forgetting that only the component of force in the direction of motion does work. When force is applied at an angle, you must use the cosine of that angle to find the effective force component. Remember: work problems in healthcare settings often involve angled forces, so always check whether the force and motion are perfectly aligned before calculating.

Question 6

An orderly pushes a 50 kg laundry cart with a net force of 20 N over a distance of 10 m. The cart starts from rest. What is the work done by the orderly, and what is the final kinetic energy of the cart?

  1. Work = 200 J; Kinetic Energy = 200 J (correct answer)
  2. Work = 200 J; Kinetic Energy = 5000 J
  3. Work = 500 J; Kinetic Energy = 200 J
  4. Work = 500 J; Kinetic Energy = 500 J
Explanation: Physics problems involving force, motion, and energy require you to identify which formulas apply and use the work-energy theorem. When you see questions about work and kinetic energy together, remember that work equals the change in kinetic energy for an object. First, calculate the work done using W=F×dW = F \times d, where force is 20 N and distance is 10 m. This gives us W=20×10=200 JW = 20 \times 10 = 200 \text{ J}. Next, apply the work-energy theorem: the work done on an object equals its change in kinetic energy. Since the cart starts from rest (initial kinetic energy = 0), all the work done becomes the final kinetic energy. Therefore, the final kinetic energy is also 200 J. Looking at the wrong answers: Answer B incorrectly calculates kinetic energy as 5000 J, which appears to come from misusing the kinetic energy formula KE=12mv2KE = \frac{1}{2}mv^2 without first finding the correct velocity. Answer C shows work as 500 J, which might result from incorrectly multiplying mass by distance (50 × 10) instead of using force × distance. Answer D makes both errors, showing 500 J for work and claiming kinetic energy equals work when work was miscalculated. The correct answer is A: Work = 200 J; Kinetic Energy = 200 J. For HESI physics questions, always identify what's given versus what you need to find, then select the appropriate formula. Remember that work and change in kinetic energy are equivalent when no other forces (like friction) are mentioned.

Question 7

An electric hospital bed is raised, increasing its gravitational potential energy by 2,100 J. If the bed is then lowered back to its original position, what is the work done by the force of gravity on the bed during the lowering process?

  1. -2,100 J
  2. 0 J
  3. 2,100 J (correct answer)
  4. 4,200 J
Explanation: When you encounter work and energy problems, focus on the relationship between gravitational potential energy and the work done by gravity. These quantities are directly related but have opposite signs depending on the direction of motion. When the bed was raised, it gained 2,100 J of gravitational potential energy. This energy came from the motor doing positive work against gravity. Now, as the bed lowers back to its original position, gravity does positive work on the bed, converting that stored potential energy back into kinetic energy (and eventually heat when the bed stops). The work done by gravity equals the decrease in gravitational potential energy, which is 2,100 J. This makes C correct. Option A (-2,100 J) represents a common sign error. Students often think that because the bed is moving downward, the work must be negative. However, work is positive when the force and displacement are in the same direction. Gravity points downward, and the bed moves downward, so gravity does positive work. Option B (0 J) incorrectly assumes that since the bed returns to its starting position, no net work was done. While the net work by all forces combined might be zero (if the bed starts and ends at rest), gravity specifically does positive work during the descent. Option D (4,200 J) likely comes from doubling the energy value, perhaps confusing the total energy change in the round trip with the work done by gravity in one direction. Remember: Work done by gravity is positive when objects move in gravity's direction (downward) and equals the magnitude of potential energy change.

Question 8

A spring in a piece of rehabilitation equipment stores 20 J of elastic potential energy when it is compressed by 0.1 meters. If the spring is compressed further to a total distance of 0.2 meters, what is its new stored elastic potential energy?

  1. 20 J
  2. 40 J
  3. 80 J (correct answer)
  4. 160 J
Explanation: When you encounter questions about elastic potential energy in springs, remember that this energy follows a quadratic relationship with displacement - meaning small changes in compression can lead to large changes in stored energy. The elastic potential energy stored in a spring is given by PE=12kx2PE = \frac{1}{2}kx^2, where k is the spring constant and x is the compression distance. Notice that energy is proportional to the square of the displacement, not just the displacement itself. First, let's find the spring constant using the initial conditions. With 20 J stored at 0.1 m compression: 20=12k(0.1)220 = \frac{1}{2}k(0.1)^2, which gives us k=4000 N/mk = 4000 \text{ N/m}. Now, when compressed to 0.2 m: PE=12(4000)(0.2)2=2000(0.04)=80 JPE = \frac{1}{2}(4000)(0.2)^2 = 2000(0.04) = 80 \text{ J}. Alternatively, since energy scales with the square of displacement, doubling the compression (0.1 to 0.2 m) increases energy by a factor of 22=42^2 = 4, so 20×4=80 J20 × 4 = 80 \text{ J}. Choice A (20 J) incorrectly assumes energy stays constant regardless of compression. Choice B (40 J) represents the common misconception that energy scales linearly with displacement - doubling compression would double energy. Choice D (160 J) might result from incorrectly calculating (0.2/0.1)3(0.2/0.1)^3 instead of squared. Remember this key pattern: elastic potential energy increases with the square of displacement. When compression doubles, energy quadruples. This quadratic relationship appears frequently in physics problems on standardized exams.

Question 9

A gurney with a total mass of 80 kg is moving at 3 m/s. A heavy patient transfer device with a mass of 320 kg is moving at 1.5 m/s. How does the kinetic energy (KE) of the gurney compare to that of the transfer device?

  1. The gurney has half the KE of the device.
  2. The gurney has the same KE as the device. (correct answer)
  3. The gurney has double the KE of the device.
  4. The gurney has four times the KE of the device.
Explanation: When you encounter physics problems involving moving objects in healthcare settings, you're working with kinetic energy, which depends on both mass and velocity. The key insight is that kinetic energy increases with the square of velocity, making speed changes more impactful than mass changes. To find each object's kinetic energy, use the formula KE=12mv2KE = \frac{1}{2}mv^2. For the gurney: KE=12(80)(3)2=12(80)(9)=360 JKE = \frac{1}{2}(80)(3)^2 = \frac{1}{2}(80)(9) = 360 \text{ J}. For the transfer device: KE=12(320)(1.5)2=12(320)(2.25)=360 JKE = \frac{1}{2}(320)(1.5)^2 = \frac{1}{2}(320)(2.25) = 360 \text{ J}. Both objects have identical kinetic energies of 360 joules. Looking at the wrong answers: Choice A suggests the gurney has half the kinetic energy, which would be 180 J, but our calculation shows 360 J. Choice C claims the gurney has double the kinetic energy (720 J), which is incorrect. Choice D proposes four times the kinetic energy (1440 J), also wrong. These distractors likely appeal to students who might assume the heavier object automatically has more kinetic energy without considering the velocity squared term. Choice B correctly states that both objects have the same kinetic energy, despite their different masses and velocities. Remember that kinetic energy problems often contain this type of "compensation" where higher mass is offset by lower velocity. Always calculate both values completely rather than making assumptions based on mass alone—the velocity squared term frequently creates surprising results.

Question 10

A 60 kg patient is exercising on a machine that requires them to lift the 60 kg mass a vertical distance of 0.4 meters with each repetition. If the patient completes 25 repetitions in one minute, what is their average power output against gravity? (Use g ≈ 10 m/s²)

  1. 100 W (correct answer)
  2. 240 W
  3. 600 W
  4. 14,400 W
Explanation: Power problems in physics require you to understand the relationship between work, energy, and time. When you see a question involving lifting mass against gravity over time, you're dealing with gravitational power calculations. To find power output, you need to calculate the work done per unit time. Work against gravity equals the gravitational potential energy gained: W=mghW = mgh, where m is mass, g is gravitational acceleration, and h is height. Here, work per repetition = 60 kg×10 m/s2×0.4 m=240 J60 \text{ kg} \times 10 \text{ m/s}^2 \times 0.4 \text{ m} = 240 \text{ J} For 25 repetitions, total work = 25×240=6000 J25 \times 240 = 6000 \text{ J} Power equals work divided by time: P=6000 J60 s=100 WP = \frac{6000 \text{ J}}{60 \text{ s}} = 100 \text{ W} Looking at the wrong answers: Answer B (240 W) represents the work done in just one repetition, not the power over one minute. Answer C (600 W) appears to come from miscalculating either the number of repetitions or the time conversion. Answer D (14,400 W) likely results from multiplying instead of dividing by time, or confusing the total energy with power. The key trap here is confusing work with power. Work tells you total energy expended, while power tells you the rate of energy expenditure. On the HESI, always check your units and remember that power problems require dividing work by time. Practice identifying whether questions ask for work (measured in joules) or power (measured in watts).

Question 11

A medical device with a mass of 5 kg is on a shelf 2 meters high. A technician lowers it to a workbench 0.8 meters high. What is the work done by gravity on the device during this process? (Use g ≈ 10 m/s²)

  1. -60 J
  2. -40 J
  3. 40 J
  4. 60 J (correct answer)
Explanation: When you encounter physics problems involving objects moving vertically, focus on the work-energy relationship and gravitational potential energy changes. Work done by gravity depends on the vertical displacement and always acts downward. To find the work done by gravity, use W=mghW = mgh, where the height (h) is the vertical displacement. The device moves from 2.0 m to 0.8 m, so the displacement is h=2.0−0.8=1.2 mh = 2.0 - 0.8 = 1.2 \text{ m} downward. Since gravity acts downward and the object moves downward, gravity does positive work: W=mgh=(5 kg)(10 m/s2)(1.2 m)=60 JW = mgh = (5 \text{ kg})(10 \text{ m/s}^2)(1.2 \text{ m}) = 60 \text{ J} Choice A (-60 J) incorrectly applies a negative sign, likely from confusing the work done by gravity with the work done against gravity. Choice B (-40 J) makes the same sign error while also using the wrong height—probably using just the final height (0.8 m) instead of the displacement. Choice C (40 J) has the correct positive sign but uses the wrong displacement, again likely using 0.8 m instead of 1.2 m. Remember this key principle: gravity always does positive work when an object moves downward and negative work when an object moves upward, regardless of other forces involved. For HESI physics questions, always identify the direction of displacement relative to the force direction—this determines whether work is positive or negative.

Question 12

A 40 kg crash cart, initially at rest, is pushed by a net force, resulting in 500 Joules of net work being done on it. What is the final speed of the crash cart, assuming negligible friction?

  1. 5.0 m/s (correct answer)
  2. 6.25 m/s
  3. 12.5 m/s
  4. 25.0 m/s
Explanation: This problem tests your understanding of the work-energy theorem, which connects force, motion, and energy. When you see work being done on an object, think about how that energy changes the object's motion. The work-energy theorem states that net work done on an object equals its change in kinetic energy: W=ΔKE=KEfinal−KEinitialW = \Delta KE = KE_{final} - KE_{initial}. Since the crash cart starts at rest, its initial kinetic energy is zero, so all 500 J of work becomes final kinetic energy. Setting up the equation: 500 J=12mv2−0500 \text{ J} = \frac{1}{2}mv^2 - 0 Solving for velocity: 500=12(40)(v2)500 = \frac{1}{2}(40)(v^2) 500=20v2500 = 20v^2 v2=25v^2 = 25 v=5.0 m/sv = 5.0 \text{ m/s} This confirms answer A is correct. Looking at the wrong answers: B (6.25 m/s) results from incorrectly using v=Wmv = \sqrt{\frac{W}{m}} instead of the proper kinetic energy formula. C (12.5 m/s) comes from setting W=mv2W = mv^2 (forgetting the 12\frac{1}{2} factor in kinetic energy). D (25.0 m/s) results from directly taking v2=25v^2 = 25 but then using 25 as the final answer instead of taking the square root. Study tip: Always remember that kinetic energy is 12mv2\frac{1}{2}mv^2, not just mv2mv^2. That factor of 12\frac{1}{2} is crucial in work-energy problems and frequently appears as a trap in answer choices.

Question 13

A hospital elevator motor must do 400,000 J of work to lift a fully loaded car. If the motor has a power output of 20,000 W, what is the minimum time required for the lift?

  1. 0.05 s
  2. 5 s
  3. 20 s (correct answer)
  4. 8,000,000 s
Explanation: When you encounter physics problems involving work, power, and time, remember that these three quantities are connected by a fundamental relationship: power equals work divided by time, or P=WtP = \frac{W}{t}. To find the minimum time required, you need to rearrange this formula to solve for time: t=WPt = \frac{W}{P}. Substituting the given values: t=400,000 J20,000 W=20 st = \frac{400,000 \text{ J}}{20,000 \text{ W}} = 20 \text{ s}. This confirms that answer C is correct. Let's examine why the other options are wrong. Answer A (0.05 s) results from incorrectly dividing power by work instead of work by power - this represents a common algebraic error when rearranging formulas. Answer B (5 s) comes from miscalculating the division, perhaps by dropping a zero or making an arithmetic mistake with the large numbers. Answer D (8,000,000 s) results from multiplying work and power instead of dividing them, which shows a fundamental misunderstanding of the power formula. For HESI physics problems, always start by identifying the relevant formula and carefully rearranging it to solve for the unknown variable. Double-check your arithmetic when working with large numbers, and remember that power represents the rate at which work is done - higher power means the same work gets completed faster, so time should decrease as power increases.

Question 14

A heart defibrillator delivers 360 Joules of energy to a patient's chest in 10 milliseconds. Which value best represents the power delivered? (1 second = 1000 milliseconds)

  1. 3.6 W
  2. 36 W
  3. 3,600 W
  4. 36,000 W (correct answer)
Explanation: When you encounter physics problems in healthcare contexts, you're applying fundamental formulas to medical equipment. Power calculations require the formula: P=EtP = \frac{E}{t}, where power (P) is energy (E) divided by time (t). First, convert the time units to match standard SI units. Since 1 second = 1000 milliseconds, then 10 milliseconds = 0.01 seconds. Now you can calculate: P=360 J0.01 s=36,000 WP = \frac{360 \text{ J}}{0.01 \text{ s}} = 36,000 \text{ W} Option A (3.6 W) results from incorrectly dividing 360 by 100 instead of converting milliseconds properly. This represents a unit conversion error where you might have confused the relationship between milliseconds and seconds. Option B (36 W) occurs when you divide 360 J by 10 ms without converting to seconds first, essentially treating milliseconds as if they were seconds. This is a common trap when students rush through unit conversions. Option C (3,600 W) happens when you convert milliseconds incorrectly, perhaps using 0.1 seconds instead of 0.01 seconds as your time value. Option D (36,000 W) is correct because it properly converts 10 milliseconds to 0.01 seconds before applying the power formula. Study tip: On the HESI, medical equipment problems often involve unit conversions paired with basic physics formulas. Always convert all units to standard SI units (seconds, not milliseconds) before plugging values into formulas. Write out your unit conversion first to avoid calculation errors.

Question 15

A hospital bed is pushed 5 meters across a floor by a constant 300 N force. The force of friction opposing the motion is 220 N. What is the net work done on the bed?

  1. 80 J
  2. 400 J (correct answer)
  3. 1500 J
  4. 2600 J
Explanation: Work problems in physics require you to consider all forces acting on an object and understand that work equals force times distance. When multiple forces act on an object, you need the net force to calculate the work done on the system. To find the net work, start by identifying all forces. The applied force is 300 N pushing the bed forward, while friction opposes motion with 220 N. The net force is: Fnet=300 N−220 N=80 NF_{net} = 300\text{ N} - 220\text{ N} = 80\text{ N} Now apply the work formula: W=F×d=80 N×5 m=400 JW = F \times d = 80\text{ N} \times 5\text{ m} = 400\text{ J} This confirms answer B is correct. Let's examine why the other choices are wrong. Choice A (80 J) represents a common error where students calculate the net force correctly (80 N) but forget to multiply by distance, giving just the force value. Choice C (1500 J) occurs when students use only the applied force and ignore friction entirely: 300 N×5 m=1500 J300\text{ N} \times 5\text{ m} = 1500\text{ J}. Choice D (2600 J) results from incorrectly adding the forces instead of finding the net force: (300+220)×5=2600 J(300 + 220) \times 5 = 2600\text{ J}. The key strategy for work problems is to always identify whether you're asked for net work or work by individual forces. When the question asks for "net work done on" an object, you must account for all forces acting on it. Remember: opposing forces subtract from applied forces, and work always requires multiplying the relevant force by the distance moved.

Question 16

The unit of power, the Watt, is equivalent to which of the following combinations of fundamental units?

  1. Joule per meter (J/m)
  2. Joule per second (J/s) (correct answer)
  3. Kilogram-meter per second (kg·m/s)
  4. Newton per second (N/s)
Explanation: When you encounter questions about units of measurement, think about the fundamental relationship between the quantity being measured and its basic components. Power measures how quickly energy is transferred or work is done over time. Power is defined as the rate of energy transfer, which means energy divided by time. Since the Joule (J) is the unit of energy, power becomes Joules per second (J/s). You can verify this by thinking about common examples: a 60-watt light bulb consumes 60 Joules of energy every second. This makes answer choice B correct. Let's examine why the other options are incorrect. Choice A (J/m) represents energy per unit distance, which describes quantities like force or energy density, not power. Choice C (kg·m/s) represents momentum, which is mass times velocity and has nothing to do with energy transfer rates. Choice D (N/s) represents the rate of change of force over time, which relates to concepts like impulse, not power. A helpful way to remember this is that power always involves a time component in the denominator because it measures "how fast" something happens energetically. The watt is specifically designed to quantify energy flow rates - whether it's electrical power in circuits, mechanical power in engines, or metabolic power in biological systems. For the HESI exam, memorize that power equals energy divided by time, so watts always equal joules per second. This fundamental relationship appears frequently in physics problems involving energy systems.

Question 17

A motorized lift is raising a 90 kg patient at a constant velocity of 0.5 m/s. What is the power output of the lift to counteract gravity? (Use g ≈ 10 m/s²)

  1. 45 W
  2. 450 W (correct answer)
  3. 900 W
  4. 1800 W
Explanation: When you encounter physics problems involving lifting objects at constant velocity, you're dealing with power calculations where the key insight is that constant velocity means zero acceleration and balanced forces. To find the power needed to counteract gravity, you need to calculate the gravitational force first, then apply the power formula. The gravitational force on the patient is F=mg=90 kg×10 m/s2=900 NF = mg = 90 \text{ kg} \times 10 \text{ m/s}^2 = 900 \text{ N}. Since the lift moves at constant velocity, the upward force must exactly equal this gravitational force to maintain equilibrium. Power is the rate of doing work, calculated as P=F×vP = F \times v, where F is the force and v is the velocity. Therefore: P=900 N×0.5 m/s=450 WP = 900 \text{ N} \times 0.5 \text{ m/s} = 450 \text{ W}. This confirms answer B is correct. Looking at the wrong answers: A) 45 W represents a calculation error where someone might have incorrectly used 9 kg instead of 90 kg for the mass. C) 900 W is the gravitational force value in Newtons, but this ignores the velocity component entirely—a common mistake of confusing force with power. D) 1800 W suggests someone doubled the correct answer, possibly by incorrectly adding forces or making an arithmetic error in the velocity calculation. Remember for HESI physics problems: when objects move at constant velocity, forces are balanced, and power equals force times velocity. Always check your units—power should be in watts, not newtons.

Question 18

Which of the following physical quantities represents the rate at which work is done or energy is transferred?

  1. Force
  2. Work
  3. Energy
  4. Power (correct answer)
Explanation: When you encounter physics questions about rates and transfers, focus on identifying what's happening per unit time versus what's being measured as a total quantity. Power is defined as the rate at which work is done or energy is transferred. Mathematically, power equals work divided by time: P=WtP = \frac{W}{t} or energy divided by time: P=EtP = \frac{E}{t}. This means power tells you how quickly energy changes hands or how fast work gets accomplished. The unit for power is the watt, which equals one joule per second. Let's examine why the other options don't fit. Force (A) measures the push or pull acting on an object, measured in newtons, but doesn't involve a time component or rate. Work (B) represents the total energy transferred when a force moves an object through a distance (W=F×dW = F \times d), but this is the total amount, not the rate. Energy (C) is the capacity to do work, measured in joules, again representing a total quantity rather than a rate of transfer. The key distinction is that power specifically describes how fast something happens, while force, work, and energy describe what is happening or how much is involved. Think of it this way: if energy is like the total amount of water in a bucket, power is like how fast you're pouring it out. Remember this pattern: whenever you see "rate of" in a physics question, you're likely looking for a quantity that involves time in the denominator. Power is the rate quantity for both work and energy transfer.

Question 19

Which of the following scenarios best describes an object having its potential energy decrease while its kinetic energy increases?

  1. A medical supply box being lifted from the floor onto a shelf at a constant speed.
  2. A wheelchair rolling to a stop due to friction on a level hallway.
  3. An IV bag falling from its hook towards the floor. (correct answer)
  4. A physical therapy ball being compressed against a wall.
Explanation: When analyzing energy transformations, focus on how potential energy (stored energy due to position) and kinetic energy (energy of motion) change as objects move under the influence of forces like gravity. Option C correctly demonstrates the energy transformation you're looking for. As an IV bag falls from its hook, it loses height, which directly decreases its gravitational potential energy. Simultaneously, gravity accelerates the bag downward, increasing its speed and therefore its kinetic energy. This represents a classic conversion where potential energy transforms into kinetic energy. Let's examine why the other options don't fit: Option A shows a supply box being lifted at constant speed, which increases potential energy while kinetic energy remains zero (constant speed means no acceleration). Option B describes a wheelchair rolling to a stop - here kinetic energy decreases due to friction, but potential energy remains constant since it's moving on a level surface. Option D involves compressing a therapy ball against a wall, which stores elastic potential energy in the compressed ball while kinetic energy becomes zero when the ball stops moving. The key distinction is that only option C shows an object in free fall or similar motion where gravity does positive work, converting stored positional energy into energy of motion. Remember this pattern: when objects fall freely under gravity's influence, potential energy always decreases while kinetic energy increases. This fundamental energy conversion appears frequently in physics problems and represents one of the most important examples of energy transformation in everyday situations.

Question 20

A patient is moving in a motorized wheelchair at a speed of 2 m/s. If the motor's output is increased so that the wheelchair's speed becomes 4 m/s, by what factor does its kinetic energy increase?

  1. It increases by a factor of 2.
  2. It increases by a factor of 4. (correct answer)
  3. It increases by a factor of 8.
  4. It does not change if the mass is constant.
Explanation: When you encounter physics problems involving motion and energy changes, focus on the mathematical relationships between variables. Kinetic energy follows a specific formula that creates predictable patterns when speeds change. Kinetic energy is calculated using KE=12mv2KE = \frac{1}{2}mv^2, where m is mass and v is velocity. Notice that velocity is squared in this equation—this is crucial for solving this problem correctly. Let's calculate both scenarios. Initially: KE1=12m(2)2=2mKE_1 = \frac{1}{2}m(2)^2 = 2m. After the speed increase: KE2=12m(4)2=8mKE_2 = \frac{1}{2}m(4)^2 = 8m. To find the factor of increase, divide the final energy by the initial energy: 8m2m=4\frac{8m}{2m} = 4. The kinetic energy increases by a factor of 4, confirming answer B. Looking at the wrong answers: A suggests the energy doubles because the speed doubles, but this ignores the squared relationship in the kinetic energy formula. C would be correct if the speed had doubled twice (2→4→8), but we only have one doubling here. D incorrectly assumes that constant mass means no energy change, missing that velocity changes dramatically affect kinetic energy even when mass stays the same. Remember this key pattern: when speed doubles, kinetic energy quadruples due to the v2v^2 term. More generally, if speed increases by factor n, kinetic energy increases by factor n2n^2. This relationship appears frequently in physics problems and helps you quickly identify the correct answer without lengthy calculations.