IB Physics Quiz: Understand Work Energy And Power

20 questions · exam conditions

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Understand Work Energy And PowerQuestion 1 of 20

A variable force $F = 3x + 5$ (where $F$ is in Newtons and $x$ is in meters) acts on an object as it moves from $x = 2\text{ m}$ to $x = 6\text{ m}$ . What is the work done by this force?

56\text{ J}

44\text{ J}

68\text{ J}

32\text{ J}

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IB Physics Quiz

IB Physics Quiz: Understand Work Energy And Power

Practice Understand 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 Understand 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

A variable force $F = 3x + 5$ (where $F$ is in Newtons and $x$ is in meters) acts on an object as it moves from $x = 2\text{ m}$ to $x = 6\text{ m}$ . What is the work done by this force?

$56\text{ J}$
$44\text{ J}$
$68\text{ J}$ (correct answer)
$32\text{ J}$

Explanation: When you encounter a variable force problem, you need to use calculus because the standard formula

W = F \cdot d

only works for constant forces. For variable forces, work is calculated using the integral:

W = \int_{x_1}^{x_2} F(x) \, dx

. Given

F = 3x + 5

, you need to integrate from

x = 2

m to

x = 6

W = \int_2^6 (3x + 5) \, dx

The antiderivative of

3x + 5

\frac{3x^2}{2} + 5x

. Evaluating the definite integral:

W = \left[\frac{3x^2}{2} + 5x\right]_2^6

W = \left(\frac{3(6)^2}{2} + 5(6)\right) - \left(\frac{3(2)^2}{2} + 5(2)\right)

W = \left(\frac{108}{2} + 30\right) - \left(\frac{12}{2} + 10\right)

W = (54 + 30) - (6 + 10) = 84 - 16 = 68 \text{ J}

Answer A (

56

J) likely comes from calculation errors in the integration or evaluation. Answer B (

44

J) might result from incorrectly using the average force multiplied by distance, or arithmetic mistakes. Answer D (

32

J) could stem from using only the constant term (

5

) in the force equation or major computational errors. Remember: whenever force varies with position, distance, or time, you must use integration to find work. Don't fall into the trap of using

W = F \cdot d

with variable forces—it will always give you the wrong answer.

Question 2

An electric motor that is 75% efficient has a useful mechanical power output of 300 W. At what rate is heat dissipated by the motor?

75 W
100 W (correct answer)
300 W
400 W

Explanation: Efficiency

\eta = P_{out} / P_{in}

. The electrical power input is

P_{in} = P_{out} / \eta = 300\ \text{W} / 0.75 = 400\ \text{W}

. The rate of heat dissipation is the wasted power, which is the difference between the power input and the useful power output:

P_{wasted} = P_{in} - P_{out} = 400\ \text{W} - 300\ \text{W} = 100\ \text{W}

. This wasted power is primarily converted into thermal energy (heat).

Question 3

A fossil fuel power plant has three stages of energy conversion. The boiler has an efficiency of 80%, the turbine has an efficiency of 50%, and the generator has an efficiency of 95%. What is the overall efficiency of the plant in converting the chemical energy of the fuel into electrical energy?

38% (correct answer)
75%
98%
225%

Explanation: The overall efficiency of a multi-stage process is the product of the efficiencies of the individual stages. The efficiencies must be expressed as decimals for multiplication. Overall efficiency

\eta_{total} = \eta_1 \times \eta_2 \times \eta_3 = 0.80 \times 0.50 \times 0.95 = 0.38

. Expressed as a percentage, this is 38%.

Question 4

A 5.0 kg block is released from rest at the top of a 4.0 m long ramp inclined at 30° to the horizontal. It experiences a constant frictional force of 8.0 N. What is the speed of the block at the bottom of the ramp? Use $g = 10\ \text{m s}^{-2}$ .

8.1 m s⁻¹
6.3 m s⁻¹
7.5 m s⁻¹
5.2 m s⁻¹ (correct answer)

Explanation: The initial energy is purely gravitational potential energy. The vertical height is

h = L \sin\theta = 4.0 \times \sin(30^\circ) = 2.0\ \text{m}

. Initial GPE =

mgh = 5.0 \times 10 \times 2.0 = 100\ \text{J}

. As the block slides down, energy is lost to work done by friction,

W_f = F_f d = 8.0\ \text{N} \times 4.0\ \text{m} = 32\ \text{J}

. By the work-energy principle, the final kinetic energy is the initial GPE minus the work done by friction:

E_k = 100\ \text{J} - 32\ \text{J} = 68\ \text{J}

. The final speed is found from

E_k = \frac{1}{2}mv^2

. So,

68 = \frac{1}{2}(5.0)v^2

, which gives

v^2 = 27.2

and

v = \sqrt{27.2} \approx 5.2\ \text{m s}^{-1}

Question 5

An object moves 5.0 m along a straight line. For the first 2.0 m, a constant force of 10 N is applied in the direction of motion. For the next 3.0 m, the applied force is reduced to 4.0 N, still in the direction of motion. What is the total work done by the applied force?

70 J
35 J
38 J
32 J (correct answer)

Explanation: Work done is calculated for each segment of the motion and then summed. For the first segment,

W_1 = F_1 d_1 = (10\ \text{N})(2.0\ \text{m}) = 20\ \text{J}

. For the second segment,

W_2 = F_2 d_2 = (4.0\ \text{N})(3.0\ \text{m}) = 12\ \text{J}

. The total work done is the sum of the work done in each segment:

W_{total} = W_1 + W_2 = 20\ \text{J} + 12\ \text{J} = 32\ \text{J}

Question 6

A water pump lifts water at a rate of 0.50 m³ per minute from a well. The water is lifted 10 m and ejected with a speed of 4.0 m s⁻¹. What is the minimum useful power output of the pump? The density of water is 1000 kg m⁻³. Use $g = 10\ \text{m s}^{-2}$ .

83 W
833 W
900 W (correct answer)
50.1 kW

Explanation: The pump must provide power to increase both the gravitational potential energy and the kinetic energy of the water. First, find the mass flow rate in kg s⁻¹:

\dot{m} = \rho \times \text{volume rate} = 1000\ \text{kg m}^{-3} \times (0.50\ \text{m}^3 / 60\ \text{s}) = 25/3\ \text{kg s}^{-1}

. The power to increase GPE is

P_{GPE} = \dot{m}gh = (25/3)(10)(10) = 2500/3\ \text{W}

. The power to increase KE is

P_{KE} = \frac{1}{2}\dot{m}v^2 = \frac{1}{2}(25/3)(4.0)^2 = \frac{1}{2}(25/3)(16) = 200/3\ \text{W}

. The total minimum power output is the sum:

P_{total} = P_{GPE} + P_{KE} = 2500/3 + 200/3 = 2700/3 = 900\ \text{W}

Question 7

A 0.50 kg block is placed against a horizontal spring of constant $k = 800\ \text{N m}^{-1}$ which has been compressed by 0.20 m. The spring is released, and the block slides across a frictionless horizontal surface and then up a frictionless ramp. What is the maximum vertical height $h$ the block reaches on the ramp? Use $g=10\ \text{m s}^{-2}$ .

1.6 m
3.2 m (correct answer)
6.4 m
16 m

Explanation: This problem uses the conservation of energy. The initial energy is stored as elastic potential energy in the spring. This is converted into kinetic energy and then into gravitational potential energy. The initial elastic potential energy is

E_H = \frac{1}{2}k\Delta x^2 = \frac{1}{2}(800)(0.20)^2 = 400 \times 0.04 = 16\ \text{J}

. At the maximum height on the ramp, all of this energy will have been converted to gravitational potential energy,

E_p = mgh

. So,

mgh = 16\ \text{J}

h = 16 / (mg) = 16 / (0.50 \times 10) = 16 / 5 = 3.2\ \text{m}

Question 8

An object starts from rest. The net work done on the object is $W$ . After this work is done, the object has momentum $p$ . What is the net work required to be done on the object, starting from rest, to give it a momentum of $2p$ ?

$\sqrt{2}W$
$2W$
$4W$ (correct answer)
$8W$

Explanation: By the work-energy theorem, the net work done on an object equals its change in kinetic energy,

W = \Delta E_k

. Since the object starts from rest,

W = E_k

. The kinetic energy can be expressed in terms of momentum

p

and mass

m

E_k = p^2 / (2m)

. So,

W = p^2 / (2m)

. We want to find the new work

W'

required to achieve a momentum of

p' = 2p

. The new kinetic energy will be

E'_k = (p')^2 / (2m) = (2p)^2 / (2m) = 4p^2 / (2m) = 4(p^2 / (2m))

. Since

W = p^2 / (2m)

, the new work is

W' = 4W

Question 9

A 10 kg crate is lifted vertically by a rope at a constant velocity of 2.0 m s⁻¹ through a height of 5.0 m. What is the net work done on the crate during this process? Use $g \approx 10\ \text{m s}^{-2}$ .

-500 J
0 J (correct answer)
500 J
1000 J

Explanation: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy (

W_{net} = \Delta E_k

). Since the crate is lifted at a constant velocity, its speed does not change, and therefore its kinetic energy does not change (

\Delta E_k = 0

). Consequently, the net work done on the crate is zero. The positive work done by the rope (500 J) is exactly balanced by the negative work done by gravity (–500 J).

Question 10

Particle X has mass $m$ and particle Y has mass (4m). Both particles have the same non-zero kinetic energy. What is the ratio of the magnitude of the momentum of Y to the magnitude of the momentum of X ( $p_Y/p_X$ )?

1/2
1
2 (correct answer)
4

Explanation: The relationship between kinetic energy

E_k

and momentum

p

E_k = p^2 / (2m)

. Rearranging for momentum gives

p = \sqrt{2mE_k}

. Since both particles have the same kinetic energy

E_k

, the momentum is proportional to the square root of the mass (

p \propto \sqrt{m}

). The ratio of the momenta is therefore

p_Y/p_X = \sqrt{m_Y} / \sqrt{m_X} = \sqrt{4m} / \sqrt{m} = \sqrt{4} = 2

Question 11

A projectile of mass $m$ is launched with an initial kinetic energy $E$ . At the highest point of its trajectory, its kinetic energy is $E/4$ . Air resistance is negligible. What was the launch angle with respect to the horizontal?

30°
45°
60° (correct answer)
75°

Explanation: Let the launch velocity be

v

at an angle

\theta

. The initial kinetic energy is

E = \frac{1}{2}mv^2

. The velocity has components

v_x = v\cos\theta

and

v_y = v\sin\theta

. At the highest point of the trajectory, the vertical component of velocity is zero, but the horizontal component

v_x

remains unchanged. The kinetic energy at the highest point is

E_{top} = \frac{1}{2}m v_x^2 = \frac{1}{2}m(v\cos\theta)^2 = (\frac{1}{2}mv^2)\cos^2\theta = E\cos^2\theta

. We are given that

E_{top} = E/4

. Therefore,

E\cos^2\theta = E/4

, which simplifies to

\cos^2\theta = 1/4

. Taking the square root gives

\cos\theta = 1/2

. The angle

\theta

is 60°.

Question 12

A box is pushed by a person around a square horizontal path of side length 5 m, returning to its starting point. A constant kinetic friction force of 10 N opposes the motion. What is the net work done by the friction force on the box?

–200 J (correct answer)
–50 J
0 J
200 J

Explanation: Work done by friction is given by

W_f = -F_f \times d

, where

d

is the total distance travelled. The total distance is the perimeter of the square, which is

4 \times 5\ \text{m} = 20\ \text{m}

. The work done by friction is therefore

W_f = -10\ \text{N} \times 20\ \text{m} = -200\ \text{J}

. The negative sign indicates that the force opposes the displacement.

Question 13

A vehicle travels at a constant speed $v$ against a total resistive force $F$ . The engine provides power $P$ . The resistive force is proportional to the square of the speed ( $F \propto v^2$ ). What is the new power $P'$ required to travel at a constant speed of $2v$ ?

2 $P$
4 $P$
8 $P$ (correct answer)
16 $P$

Explanation: The power required to overcome a force

F

at speed

v

P = Fv

. We are given that the resistive force

F

is proportional to

v^2

, so we can write

F = kv^2

for some constant

k

. Substituting this into the power equation gives

P = (kv^2)v = kv^3

. Therefore, power is proportional to the cube of the speed. If the speed is doubled from

v

2v

, the new power

P'

will be

k(2v)^3 = k(8v^3) = 8(kv^3) = 8P

Question 14

A power station produces the same amount of useful electrical energy per day by burning either 10 tonnes of fuel A (energy density 45 MJ kg⁻¹) or 12 tonnes of fuel B (energy density 30 MJ kg⁻¹). What is the ratio of the efficiency when using Fuel A to the efficiency when using Fuel B ( $\eta_A / \eta_B$ )?

0.80 (correct answer)
0.83
1.25
1.50

Explanation: Efficiency is

\eta = E_{out} / E_{in}

. Since

E_{out}

is the same for both fuels, the ratio of efficiencies is

\eta_A / \eta_B = (E_{out}/E_{in,A}) / (E_{out}/E_{in,B}) = E_{in,B} / E_{in,A}

. First, calculate the total energy input for each fuel.

E_{in,A} = (10 \times 1000\ \text{kg}) \times (45\ \text{MJ kg}^{-1}) = 450000\ \text{MJ}

E_{in,B} = (12 \times 1000\ \text{kg}) \times (30\ \text{MJ kg}^{-1}) = 360000\ \text{MJ}

. The ratio is

\eta_A / \eta_B = 360000 / 450000 = 36/45 = 4/5 = 0.80

Question 15

An external force compresses a spring that obeys Hooke's Law by 0.20 m. The force required to hold the spring at this compression is 50 N. How much work was done by the external force to compress the spring?

5.0 J (correct answer)
10 J
25 J
50 J

Explanation: The work done in compressing a spring is equal to the elastic potential energy stored in it,

E_H = \frac{1}{2}k\Delta x^2

. First, find the spring constant

k

from Hooke's Law,

F = k\Delta x

. So,

k = F/\Delta x = 50\ \text{N} / 0.20\ \text{m} = 250\ \text{N m}^{-1}

. Now, calculate the work done:

W = \frac{1}{2}(250\ \text{N m}^{-1})(0.20\ \text{m})^2 = \frac{1}{2}(250)(0.04) = 5.0\ \text{J}

. A common mistake is to calculate work as

W = F\Delta x = 50 \times 0.20 = 10\ \text{J}

, which is incorrect because the force is not constant during the compression.

Question 16

A Sankey diagram for a heat engine shows an input energy arrow of width 10 units. An arrow representing useful work has a width of 3 units, and an arrow representing waste heat has a width of 7 units.

The engine is modified to increase its efficiency. If the energy input remains the same, how will the Sankey diagram change?

The input arrow will be narrower while the output arrows remain the same.
The useful work arrow will be wider and the waste heat arrow will be narrower. (correct answer)
The useful work arrow will be narrower and the waste heat arrow will be wider.
Both the useful work and waste heat arrows will become wider.

Explanation: Efficiency is the ratio of useful work output to total energy input. If the efficiency increases with the same energy input, a larger fraction of the input energy must be converted into useful work. In a Sankey diagram, the width of the arrow is proportional to the amount of energy. Therefore, the arrow for useful work must become wider. By the principle of conservation of energy, if the useful work output increases, the waste heat must decrease (since Input = Useful Work + Waste Heat). Thus, the waste heat arrow will become narrower.

Question 17

A car of mass $m$ starts from rest on a horizontal road and is accelerated by an engine that delivers constant power $P$ . Air resistance is negligible. Which statement best describes the acceleration of the car?

The acceleration is constant throughout the motion.
The acceleration is initially large and decreases as speed increases. (correct answer)
The acceleration is initially small and increases as speed increases.
The acceleration is zero because the power is constant.

Explanation: Power is given by

P = Fv

, where

F

is the driving force and

v

is the instantaneous speed. From Newton's second law,

F = ma

. Substituting this into the power equation gives

P = (ma)v

. Rearranging for acceleration

a

gives

a = P / (mv)

. Since

P

and

m

are constant, the acceleration

a

is inversely proportional to the speed

v

. As the car starts from rest and its speed increases, its acceleration must decrease.

Question 18

A 2.0 kg mass is moved from point X at a height of 1.0 m above the ground to point Y at a height of 3.0 m. Path 1 is a direct vertical lift. Path 2 is a 10 m long frictionless ramp. Let $W_1$ be the work done against gravity for Path 1 and $W_2$ be the work done against gravity for Path 2. Which statement is correct?

$W_1 = W_2$ (correct answer)
$W_1 < W_2$
$W_1 > W_2$
The relationship depends on the angle of the ramp.

Explanation: The gravitational force is a conservative force. This means the work done against gravity depends only on the change in vertical height (potential energy), not on the path taken. The change in height is

\Delta h = 3.0\ \text{m} - 1.0\ \text{m} = 2.0\ \text{m}

for both paths. Therefore, the work done against gravity is the same for both paths:

W = mg\Delta h

Question 19

A block slides down a frictionless inclined plane of height $h$ and then compresses a spring at the bottom. If the spring constant is $k$ and the block comes to rest after compressing the spring by distance $x$ , what is the mass of the block?

$m = \frac{kx^2}{2gh}$ (correct answer)
$m = \frac{2gh}{kx^2}$
$m = \frac{kx}{2gh}$
$m = \frac{gh}{kx^2}$

Explanation: Using conservation of energy: Initial gravitational potential energy equals final elastic potential energy.

mgh = \frac{1}{2}kx^2

. Solving for mass:

m = \frac{kx^2}{2gh}

. Choice B inverts the relationship incorrectly. Choice C omits the square on

x

. Choice D is missing the factor of 2.

Question 20

A block is pulled up a $30°$ incline by a force $F$ parallel to the incline. The coefficient of kinetic friction is $0.2$ . If the block moves at constant velocity, what fraction of the work done by force $F$ is dissipated as heat due to friction?

$0.50$
$0.26$ (correct answer)
$0.20$
$0.65$

Explanation: When analyzing work and energy on inclined planes with friction, you need to identify where the applied force's energy goes. Since the block moves at constant velocity, the net force is zero, meaning the applied force

F

exactly balances both the gravitational component down the incline and friction. Let's set up the force balance. The weight component down the incline is

mg\sin(30°) = 0.5mg

. The friction force is

\mu_k mg\cos(30°) = 0.2mg \times \frac{\sqrt{3}}{2} = 0.1\sqrt{3}mg \approx 0.173mg

. Therefore,

F = mg\sin(30°) + \mu_k mg\cos(30°) = 0.5mg + 0.173mg = 0.673mg

. When the block moves distance

d

up the incline, the work done by force

F

W_F = Fd = 0.673mgd

. The work dissipated by friction is

W_{friction} = \mu_k mg\cos(30°) \times d = 0.173mgd

. The fraction dissipated as heat is

\frac{W_{friction}}{W_F} = \frac{0.173mgd}{0.673mgd} = \frac{0.173}{0.673} = 0.26

. Answer B (0.26) is correct. Answer A (0.50) incorrectly assumes friction equals the gravitational component. Answer C (0.20) mistakenly uses just the coefficient of friction without considering the geometry. Answer D (0.65) appears to invert the calculation or use an incorrect force relationship. Remember: on inclined planes with friction, always resolve forces parallel and perpendicular to the surface, and account for both gravitational and frictional components when calculating work distribution.