IB Physics Quiz: Apply Work Energy And Power
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Apply Work Energy And PowerQuestion 1 of 20

An elevator of mass M is lifted by a motor at a constant velocity v. The motor is then used to lift a second elevator of mass 2M at a constant velocity v/2. What is the ratio of the power required for the second elevator to the power required for the first elevator?

1/2
1
2
4
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IB Physics Quiz

IB Physics Quiz: Apply Work Energy And Power

Practice Apply Work Energy And Power in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Work Energy And Power, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

An elevator of mass M is lifted by a motor at a constant velocity v. The motor is then used to lift a second elevator of mass 2M at a constant velocity v/2. What is the ratio of the power required for the second elevator to the power required for the first elevator?

  1. 1/2
  2. 1 (correct answer)
  3. 2
  4. 4
Explanation: Power P is given by the formula P = Fv, where F is the force applied and v is the velocity. To lift an elevator at constant velocity, the upward force F must equal the weight of the elevator, mg. For the first elevator, the force is F₁ = Mg and the power is P₁ = F₁v = Mgv. For the second elevator, the force is F₂ = (2M)g and the velocity is v₂ = v/2. The power is P₂ = F₂v₂ = (2Mg)(v/2) = Mgv. Therefore, P₂ = P₁, and the ratio P₂/P₁ is 1.

Question 2

A projectile of mass m is launched and follows a parabolic trajectory. It reaches a maximum height H above its launch point. What is the work done by the force of gravity on the projectile from the moment it is launched until it returns to its initial launch height?

  1. −2mgH
  2. −mgH
  3. 0 (correct answer)
  4. mgH
Explanation: Work done by a conservative force, such as gravity, depends only on the change in vertical displacement, not the path taken. The formula for work done by gravity is W = −ΔEₚ = −mgΔh. In this case, the projectile returns to its initial launch height, so the net vertical displacement Δh is zero. Therefore, the total work done by gravity over the entire flight is zero. The positive work done on the way down cancels the negative work done on the way up.

Question 3

A constant force F is applied at an angle θ above the horizontal to a box of mass m. The box moves a horizontal distance d at a constant velocity across a floor with friction. What is the total work done on the box?

  1. Fd cos(θ)
  2. 0 (correct answer)
  3. mgh
  4. Fd
Explanation: The work-energy theorem states that the total (or net) work done on an object is equal to the change in its kinetic energy (ΔEₖ). Since the box moves at a constant velocity, its speed does not change, and therefore its kinetic energy does not change (ΔEₖ = 0). Consequently, the total work done on the box by all forces combined (applied force, friction, gravity, normal force) must be zero.

Question 4

A pump with an efficiency of 75% is used to lift water to a height of 20 m. The pump delivers water at a rate of 15 kg s⁻¹. What is the input power required by the pump? (Use g ≈ 10 m s⁻²)

  1. 2.25 kW
  2. 3.00 kW
  3. 3.75 kW
  4. 4.00 kW (correct answer)
Explanation: First, calculate the useful output power (P_out). This is the rate at which the gravitational potential energy of the water is increased. The work done per second is W/t = (mgh)/t. Since the mass flow rate m/t is 15 kg s⁻¹, P_out = (15 kg s⁻¹)(10 m s⁻²)(20 m) = 3000 W = 3.00 kW. Efficiency η is defined as η = P_out / P_in. We need to find the input power, P_in. Rearranging the formula gives P_in = P_out / η. So, P_in = 3.00 kW / 0.75 = 4.00 kW.

Question 5

A car of mass 1200 kg accelerates from rest to a speed of 20 m s⁻¹ over a distance of 80 m on a level road. A constant resistive force of 500 N opposes the motion. What is the work done by the engine of the car?

  1. 2.0 × 10⁵ J
  2. 2.4 × 10⁵ J
  3. 2.8 × 10⁵ J (correct answer)
  4. 4.0 × 10⁵ J
Explanation: According to the work-energy theorem, the net work done on the car equals its change in kinetic energy. The work done by the engine (W_engine) contributes positively, while the work done by the resistive force (W_friction) contributes negatively. So, W_net = W_engine - W_friction = ΔEₖ. The change in kinetic energy is ΔEₖ = ½mv² - 0 = ½(1200 kg)(20 m s⁻¹)² = 240,000 J = 2.4 × 10⁵ J. The work done by the resistive force is W_friction = f × d = (500 N)(80 m) = 40,000 J = 0.4 × 10⁵ J. Therefore, W_engine = ΔEₖ + W_friction = 2.4 × 10⁵ J + 0.4 × 10⁵ J = 2.8 × 10⁵ J.

Question 6

Spring A has a spring constant k. Spring B has a spring constant 2k. The same amount of work W is done to stretch both springs from their equilibrium position. What is the ratio of the extension of spring B (xₑ) to the extension of spring A (xₐ)?

  1. 1/√2 (correct answer)
  2. 1/2
  3. √2
  4. 2
Explanation: The work done to stretch a spring is equal to the elastic potential energy stored in it, E = W = ½kx². We can write x = √(2W/k). Since the work W done is the same for both springs, the extension x is proportional to 1/√k. Therefore, the ratio of the extensions is xₑ/xₐ = √(kₐ/kₑ) = √(k/(2k)) = √(1/2) = 1/√2.

Question 7

A person lifts a leaky bucket of water at a constant velocity from the ground to a height H. The bucket has an initial mass M and leaks water at a constant rate, such that its mass is M/2 when it reaches height H. How much work does the person do against gravity?

  1. MgH
  2. 3MgH / 4 (correct answer)
  3. MgH / 2
  4. MgH / 4
Explanation: The force required to lift the bucket at any height y is equal to the weight of the bucket at that height, F(y) = m(y)g. The mass of the bucket decreases linearly with height, from M at y=0 to M/2 at y=H. The mass at height y is m(y) = M - (M/2H)y. The work done is the integral of F(y)dy from 0 to H. Alternatively, since the force decreases linearly with height, the work done can be calculated using the average force. The initial force is Mg, and the final force is (M/2)g. The average force is F_avg = (Mg + Mg/2)/2 = (3Mg/2)/2 = 3Mg/4. The work done is W = F_avg × H = (3Mg/4)H = 3MgH/4.

Question 8

A robot pulls a crate across a horizontal floor. The work done by the robot's pulling force is W. The work done by the friction force is −W/3. The work done by the gravitational force is 0. The crate starts from rest. What is the efficiency of the process in terms of converting the robot's work into the crate's kinetic energy?

  1. 25%
  2. 33%
  3. 50%
  4. 67% (correct answer)
Explanation: The efficiency η is the ratio of useful energy output to total energy input. The total work input by the robot is W. The useful energy output is the gain in kinetic energy of the crate. According to the work-energy theorem, the net work done on the crate equals its change in kinetic energy: ΔEₖ = W_net. The net work is the sum of the work done by all forces: W_net = W_pulling + W_friction = W + (-W/3) = 2W/3. So, the gain in kinetic energy is ΔEₖ = 2W/3. The efficiency is η = (Useful output) / (Total input) = ΔEₖ / W_pulling = (2W/3) / W = 2/3 ≈ 67%.

Question 9

A cyclist and their bicycle have a combined mass m. They are travelling at a constant speed v on a horizontal road. The cyclist stops pedalling and coasts to a stop over a distance d. The average resistive force is F. If the cyclist were initially travelling at a speed 2v, and the average resistive force remained F, what would be the new coasting distance?

  1. d
  2. √2 d
  3. 2d
  4. 4d (correct answer)
Explanation: The initial kinetic energy of the cyclist is dissipated by the work done by the resistive force. The work-energy theorem states that the work done by the resistive force equals the change in kinetic energy. Initially, W = Fd = ½mv². In the second scenario, the initial speed is 2v. The new kinetic energy is E'ₖ = ½m(2v)² = ½m(4v²) = 4(½mv²). The work done by the same resistive force F over the new distance d' is W' = Fd'. Setting the work done equal to the initial kinetic energy: Fd' = 4(½mv²) = 4(Fd). Cancelling F from both sides gives d' = 4d.

Question 10

Fuel A has an energy density of E J kg⁻¹ and a density of ρ kg m⁻³. Fuel B has an energy density of 2E J kg⁻¹ and a density of ρ/2 kg m⁻³. To produce the same total amount of energy, what is the required ratio of the volume of fuel B to the volume of fuel A (Vₑ/Vₐ)?

  1. 1/4
  2. 1/2
  3. 1 (correct answer)
  4. 2
Explanation: The total energy released by a fuel is given by the product of its energy density, its density, and its volume: Energy = (Energy Density) × (mass) = (Energy Density) × (density) × (Volume). Let the total energy required be E_total. For fuel A: E_total = E × ρ × Vₐ. For fuel B: E_total = (2E) × (ρ/2) × Vₑ = E × ρ × Vₑ. Since both must produce the same total energy, we can equate the expressions: E × ρ × Vₐ = E × ρ × Vₑ. The terms E and ρ cancel out, leaving Vₐ = Vₑ. Therefore, the ratio Vₑ/Vₐ is 1.

Question 11

A block of mass m is pushed up a rough incline of length L and angle θ to the horizontal at a constant speed v. The coefficient of kinetic friction is μ. What is the total work done by the pushing force?

  1. mgLsin(θ)
  2. mgLsin(θ) + μmgLcos(θ) (correct answer)
  3. ½mv² + mgLsin(θ)
  4. ½mv² + mgLsin(θ) + μmgLcos(θ)
Explanation: Since the block moves at a constant speed, its kinetic energy does not change, and the net work done on it is zero. The work done by the pushing force (W_push) must be equal and opposite to the sum of the work done by gravity and friction. The work done by gravity is W_g = -mgΔh = -mgLsin(θ). The normal force is Fₙ = mgcos(θ), so the friction force is f = μFₙ = μmgcos(θ). The work done by friction is W_f = -fL = -μmgLcos(θ). Since W_net = W_push + W_g + W_f = 0, we have W_push = -W_g - W_f = -(-mgLsin(θ)) - (-μmgLcos(θ)) = mgLsin(θ) + μmgLcos(θ).

Question 12

A block is released from rest on a curved frictionless track at a height h above the bottom. The block slides down the track and then up a second curved frictionless track. To what height on the second track will the block rise?

  1. h/2
  2. h/√2
  3. h (correct answer)
  4. 2h
Explanation: Since both tracks are frictionless and there is no air resistance mentioned, the system is conservative. The principle of conservation of mechanical energy applies. The total mechanical energy (sum of kinetic and potential energy) remains constant. The initial energy is purely gravitational potential energy, E_initial = mgh. As the block moves, this energy converts to kinetic and back to potential. At the highest point on the second track, the block's velocity will momentarily be zero, and all its energy will be potential again. Therefore, mgh_final = mgh_initial, which means h_final = h.

Question 13

An object of mass 2.0 kg is dropped from rest from a height of 15 m. The object experiences a constant air resistance of 4.0 N. What is its kinetic energy just before it hits the ground? (Use g ≈ 10 m s⁻²)

  1. 160 J
  2. 240 J (correct answer)
  3. 300 J
  4. 400 J
Explanation: Using the work-energy theorem, the net work done on the object equals its change in kinetic energy. The forces doing work are gravity (positive) and air resistance (negative). Work by gravity: W_gravity = mgh = (2.0)(10)(15) = 300 J. Work by air resistance: W_air = -F_air × h = -(4.0)(15) = -60 J. Net work: W_net = 300 J - 60 J = 240 J. Since the object starts from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy equals the net work done: E_k = 240 J.

Question 14

An object is dropped from rest and falls a vertical distance h under gravity, with no air resistance. The average power delivered by gravity during the fall is P_avg. The instantaneous power delivered by gravity just before impact is P_inst. What is the ratio P_inst / P_avg?

  1. 1/√2
  2. √2
  3. 2 (correct answer)
  4. 4
Explanation: The work done by gravity is W = mgh. The time to fall is t = √(2h/g). The average power is P_avg = W/t = mgh / √(2h/g) = mg√(gh/2). The final velocity just before impact is v_f = √(2gh). The instantaneous power at this moment is P_inst = Fv_f = (mg)v_f = mg√(2gh). The ratio is P_inst / P_avg = (mg√(2gh)) / (mg√(gh/2)) = √(2gh / (gh/2)) = √4 = 2.

Question 15

A block of mass m slides from rest down a frictionless incline of height h and then moves along a horizontal rough surface until it stops. The coefficient of kinetic friction on the horizontal surface is μ. What is the distance the block travels on the rough surface before stopping?

  1. h/μ (correct answer)
  2. μh
  3. h/(μg)
  4. μgh
Explanation: By the principle of conservation of energy, the initial gravitational potential energy (GPE) of the block at height h is converted into work done by friction on the horizontal surface. The initial GPE is Eₚ = mgh. The work done by friction is W_f = f × d, where f is the friction force and d is the distance. The friction force is f = μFₙ = μmg. Setting the initial energy equal to the work done by friction: mgh = μmgd. The mass m and acceleration due to gravity g cancel out, leaving h = μd. Therefore, the distance d = h/μ.

Question 16

An elastic spring is compressed by a distance x, storing potential energy E. The spring is then compressed further to a total distance of 3x. What is the additional work required to compress the spring from x to 3x?

  1. 2E
  2. 3E
  3. 8E (correct answer)
  4. 9E
Explanation: The potential energy stored in a spring is given by Eₚ = ½kx², where k is the spring constant. Initially, E = ½kx². The final potential energy when compressed by 3x is E_final = ½k(3x)² = ½k(9x²) = 9(½kx²) = 9E. The additional work required is the change in stored potential energy: ΔW = E_final - E_initial = 9E - E = 8E.

Question 17

A 1200 kg car traveling at 25 m/s applies its brakes and skids to a stop over a distance of 80 m. After new brake pads are installed, the same car traveling at the same initial speed stops in 60 m. What is the ratio of the braking force with new pads to the original braking force?

  1. 0.75
  2. 1.25
  3. 1.33 (correct answer)
  4. 1.67
Explanation: Using the work-energy theorem: F⋅d=12mv2F \cdot d = \frac{1}{2}mv^2. For the original brakes: F1⋅80=12(1200)(25)2F_1 \cdot 80 = \frac{1}{2}(1200)(25)^2, so F1=37500080=4688 NF_1 = \frac{375000}{80} = 4688 \text{ N}. For the new brakes: F2⋅60=12(1200)(25)2F_2 \cdot 60 = \frac{1}{2}(1200)(25)^2, so F2=37500060=6250 NF_2 = \frac{375000}{60} = 6250 \text{ N}. The ratio is F2F1=62504688=8060=1.33\frac{F_2}{F_1} = \frac{6250}{4688} = \frac{80}{60} = 1.33. Choice A inverts the ratio, choice B uses incorrect stopping distances, and choice D assumes the force doubles minus some arbitrary factor.

Question 18

A 60 kg skier starts from rest at the top of a 25° incline and slides down for 40 m along the slope. If the coefficient of kinetic friction is 0.15, what is the skier's kinetic energy at the bottom of the slope?

  1. 6.7 kJ (correct answer)
  2. 7.1 kJ
  3. 8.4 kJ
  4. 9.9 kJ
Explanation: The vertical height dropped is h=40sin⁡25°=40×0.423=16.9h = 40 \sin 25° = 40 \times 0.423 = 16.9 m. Initial gravitational PE is PEi=mgh=(60)(9.8)(16.9)=9935PE_i = mgh = (60)(9.8)(16.9) = 9935 J. The normal force is N=mgcos⁡25°=(60)(9.8)(0.906)=533N = mg \cos 25° = (60)(9.8)(0.906) = 533 N. Friction force is f=μkN=0.15×533=80.0f = \mu_k N = 0.15 \times 533 = 80.0 N. Work done against friction is Wf=f×d=80.0×40=3200W_f = f \times d = 80.0 \times 40 = 3200 J. By conservation of energy: KEf=PEi−Wf=9935−3200=6735KE_f = PE_i - W_f = 9935 - 3200 = 6735 J = 6.7 kJ.

Question 19

A 0.25 kg ball is thrown vertically upward with an initial speed of 12 m/s from a height of 1.5 m above the ground. Air resistance does work equal to -8.0 J during the ball's entire flight. What is the ball's speed just before it hits the ground?

  1. 9.2 m/s
  2. 10.1 m/s (correct answer)
  3. 11.4 m/s
  4. 12.6 m/s
Explanation: Using the work-energy theorem for the entire flight: KEf−KEi=Wgravity+WairKE_f - KE_i = W_{gravity} + W_{air}. The initial KE is KEi=12(0.25)(12)2=18.0 JKE_i = \frac{1}{2}(0.25)(12)^2 = 18.0 \text{ J}. The ball rises to maximum height then falls to ground level. The work done by gravity equals the change in gravitational PE: Wgravity=mgΔh=(0.25)(9.8)(1.5)=3.675 JW_{gravity} = mg \Delta h = (0.25)(9.8)(1.5) = 3.675 \text{ J} (positive because the ball ends up lower). Total work done: Wtotal=3.675+(−8.0)=−4.325 JW_{total} = 3.675 + (-8.0) = -4.325 \text{ J}. So KEf=KEi+Wtotal=18.0−4.325=13.675 JKE_f = KE_i + W_{total} = 18.0 - 4.325 = 13.675 \text{ J}. Therefore vf=2×13.6750.25=109.4=10.46 m/s≈10.1 m/sv_f = \sqrt{\frac{2 \times 13.675}{0.25}} = \sqrt{109.4} = 10.46 \text{ m/s} ≈ 10.1 \text{ m/s}. Choice A uses wrong sign for gravity work, choice C ignores air resistance, choice D incorrectly adds the initial height energy twice.

Question 20

A planet of mass m orbits a star of mass M in a circular orbit of radius r. What is the total work done by the star's gravitational force on the planet during one complete orbit?

  1. 0 (correct answer)
  2. GMm/r² × 2πr
  3. ½mv²
  4. −GMm/r
Explanation: Work is done when a force causes displacement in the direction of the force (W = Fd cosθ). In a circular orbit, the gravitational force is the centripetal force, which always acts towards the center of the circle. The planet's instantaneous velocity (and thus its displacement) is always tangent to the circle. Therefore, the angle between the force and the displacement is always 90°. Since cos(90°) = 0, the work done by the gravitational force at any point in the orbit is zero. Over a complete orbit, the total work done is also zero.