Work, Energy and Power Practice Questions — JAMB UTME

Question 1Question →

A mountain climber of mass $60\text{ kg}$ climbs a vertical height of $15\text{ m}$ in a time of $30\text{ s}$ . Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , calculate the average power expended by the climber in Watts.

Show answer & explanation

Answer: 300

Answer

The average power expended by the climber is

300\text{ W}

.

The work done in lifting a mass

m

through vertical height

h

is given by

W = mgh = 60 \times 10 \times 15 = 9000\text{ J}

. The average power is the rate of doing work,

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Step-by-Step Solution

1

Identify the given values and formula for work done against gravity.

Mass

m = 60\text{ kg}

, vertical displacement

h = 15\text{ m}

, acceleration due to gravity

g = 10\text{ m s}^{-2}

, and time

t = 30\text{ s}

. Work done formula:

W = mgh

.

The climber works against gravity to increase their potential energy by an amount equal to

mgh

.

2

Calculate the total work done.

W = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}

.

Multiplying force (

mg

) by vertical distance (

h

) yields work done in Joules.

3

Calculate the average power.

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Power is defined as work done divided by the time interval (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

Estimated Time:45s

Question 2Question →

A bullet of mass $20\text{ g}$ is moving horizontally at a speed of $400\text{ m s}^{-1}$ . It strikes a stationary block of wood and emerges from the opposite side with a speed of $100\text{ m s}^{-1}$ . What is the magnitude of the work done by the bullet against the resistance of the block, in Joules?

Show answer & explanation

Answer: 1500

Answer

The magnitude of the work done by the bullet in penetrating the block is

1500\text{ J}

.

According to the Work-Energy Theorem, the net work done on an object equals the change in its kinetic energy. The reduction in kinetic energy as the bullet slows from

400\text{ m s}^{-1}

to

100\text{ m s}^{-1}

equals the work done against the resistive force of the wooden block.

Step-by-Step Solution

1

Convert the mass of the bullet into standard SI units (kilograms)

m = \frac{20}{1000} = 0.02\text{ kg}

The standard SI unit for mass in energy equations is the kilogram.

2

Determine the initial kinetic energy of the bullet before entering the block

E_{ki} = \frac{1}{2} \times 0.02\text{ kg} \times (400\text{ m s}^{-1})^2 = 1600\text{ J}

Kinetic energy is defined as

E_k = \frac{1}{2} m v^2

.

3

Determine the final kinetic energy of the bullet as it emerges from the block

E_{kf} = \frac{1}{2} \times 0.02\text{ kg} \times (100\text{ m s}^{-1})^2 = 100\text{ J}

The bullet loses speed upon passing through the block, reducing its kinetic energy.

4

Apply the Work-Energy Theorem to find the work done against resistive forces

W = \Delta E_k = 1600\text{ J} - 100\text{ J} = 1500\text{ J}

The net work done on the bullet equals its change in kinetic energy.

Key Concept

Work-Energy Theorem

Question 3Question →

A water pump driven by an engine with an efficiency of $80\%$ raises water from an underground tank of depth $20\text{ m}$ and discharges it through a nozzle of cross-sectional area $10\text{ cm}^2$ at a steady speed of $10\text{ m/s}$ . Taking the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the minimum input power rating (in $\text{W}$ ) required for the engine?

Show answer & explanation

Answer: 3125

Answer

The minimum input power rating required for the engine is

3125\text{ W}

.

The engine must supply power to lift

10\text{ kg}

of water per second through a vertical height of

20\text{ m}

while accelerating it to

10\text{ m/s}

. The useful power output is

2000\text{ W}

(potential)

+ 500\text{ W}

(kinetic)

= 2500\text{ W}

. Accounting for an engine efficiency of

80\%

, the total input power is

\frac{2500}{0.80} = 3125\text{ W}

.

Step-by-Step Solution

1

Determine the mass of water discharged per unit time (mass flow rate).

\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}

Water is moving through a cross-sectional area at a constant velocity.

2

Calculate the useful output power required to lift the water and impart kinetic energy.

P_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}

The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.

3

Calculate the total input power using engine efficiency.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}

Efficiency is the ratio of useful power output to total power input.

Key Concept

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations

Question 4Question →

A body of mass $4\text{ kg}$ moves with a constant velocity of $5\text{ m s}^{-1}$ . Calculate the kinetic energy of the body in Joules.

Show answer & explanation

Answer: 50

Answer

The kinetic energy of the body is

50\text{ J}

.

The kinetic energy of an object in linear motion is calculated using

E_k = \frac{1}{2} m v^2

. Substituting

m = 4\text{ kg}

and

v = 5\text{ m s}^{-1}

gives

E_k = \frac{1}{2} \times 4 \times 5^2 = 50\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 4\text{ kg}

and velocity

v = 5\text{ m s}^{-1}

Extract values provided in the problem statement.

2

Apply the formula for kinetic energy

E_k = \frac{1}{2} m v^2

Kinetic energy is defined as half the product of mass and the square of velocity.

3

Substitute values and solve

E_k = \frac{1}{2} \times 4 \times (5)^2 = 50\text{ J}

Perform arithmetic evaluation to get the final energy in Joules.

Key Concept

Kinetic Energy

Question 5Question →

A block is pulled along a smooth horizontal table by a horizontal force of $30\text{ N}$ through a displacement of $4\text{ m}$ . At the same time, a vertical upward force of $40\text{ N}$ acts on the block as it moves. What is the total work done on the block?

120\text{ J}

280\text{ J}

160\text{ J}

7.5\text{ J}

Show answer & explanation

Answer: $120\text{ J}$

Answer

The total work done on the block is

120\text{ J}

.

Work done is given by

W = F d \cos\theta

. The horizontal force acts along the line of motion (

\theta = 0^\circ

), performing

30\text{ N} \times 4\text{ m} = 120\text{ J}

of work. The vertical force acts at an angle of

90^\circ

to the horizontal displacement, performing zero work because

\cos(90^\circ) = 0

. Thus, the total work done is

120\text{ J}

.

Step-by-Step Solution

1

Identify the formula for work done by a constant force

W = F \cdot d \cdot \cos(\theta)

Work is defined as the scalar product of force and displacement vectors.

2

Calculate work done by the horizontal force

W_{\text{horizontal}} = 30\text{ N} \times 4\text{ m} \times \cos(0^\circ) = 120\text{ J}

The horizontal force is in the exact direction of motion (

\theta = 0^\circ

).

3

Calculate work done by the vertical force

W_{\text{vertical}} = 40\text{ N} \times 4\text{ m} \times \cos(90^\circ) = 0\text{ J}

The vertical force is perpendicular to the horizontal displacement (

\theta = 90^\circ

), so

\cos(90^\circ) = 0

.

4

Sum the work done by all forces

W_{\text{total}} = 120\text{ J} + 0\text{ J} = 120\text{ J}

Work is a scalar quantity, so total work is the algebraic sum of individual work values.

Key Concept

Work done by perpendicular forces is zero

Estimated Time:45s

Question 6Question →

A variable force $F$ acts on a body moving along a straight horizontal path. The force increases linearly from $0\text{ N}$ at position $x = 0\text{ m}$ to $20\text{ N}$ at $x = 4\text{ m}$ , remains constant at $20\text{ N}$ from $x = 4\text{ m}$ to $x = 7\text{ m}$ , and then decreases linearly back to $0\text{ N}$ at $x = 10\text{ m}$ . What is the total work done by the force in Joules over the $10\text{ m}$ displacement?

Show answer & explanation

Answer: 130

Answer

130 J

The work done by a variable force is equal to the total area under the force-displacement (

F

-

x

) graph. The region under the graph forms a trapezoid bounded by parallel sides of length

10\text{ m}

(total displacement) and

3\text{ m}

(constant force interval), with a height of

20\text{ N}

. Calculating the area yields

W = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}

.

Step-by-Step Solution

1

Relate work done to the force-displacement graph

Work done

W

equals the total area under the

F

-

x

graph between

x = 0\text{ m}

and

x = 10\text{ m}

.

By definition,

W = \int F \, dx

, which corresponds to the geometric area under the force-displacement curve.

2

Calculate the geometric area under each section of the graph

First section (

0

to

4\text{ m}

): Triangle area =

\frac{1}{2} \times 4 \times 20 = 40\text{ J}

. Second section (

4

to

7\text{ m}

): Rectangle area =

(7 - 4) \times 20 = 60\text{ J}

. Third section (

7

to

10\text{ m}

): Triangle area =

\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}

.

Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.

3

Sum the areas of all sections

Total Work

W = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}

.

The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Key Concept

Work Done by a Variable Force (Area under Force-Displacement Graph)

Question 7Question →

A box of mass $2\text{ kg}$ slides down a rough inclined plane from a height of $5\text{ m}$ . If it reaches the bottom of the incline with a speed of $6\text{ m s}^{-1}$ , what is the work done against friction during the descent? (Take $g = 10\text{ m s}^{-2}$ )

64\text{ J}

36\text{ J}

100\text{ J}

136\text{ J}

Show answer & explanation

Answer: $64\text{ J}$

Answer

64\text{ J}

According to the principle of conservation of energy, the work done against friction equals the loss in total mechanical energy. The initial potential energy is

mgh = 2 \times 10 \times 5 = 100\text{ J}

, and the final kinetic energy is

\frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 6^2 = 36\text{ J}

. Subtracting final kinetic energy from initial potential energy yields

100\text{ J} - 36\text{ J} = 64\text{ J}

.

Step-by-Step Solution

1

Calculate the initial potential energy (

E_p

) at height

h = 5\text{ m}

.

E_p = mgh = 2\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 100\text{ J}

At the top of the incline, all mechanical energy is stored as gravitational potential energy.

2

Calculate the final kinetic energy (

E_k

) at the bottom where

v = 6\text{ m s}^{-1}

.

E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2\text{ kg} \times (6\text{ m s}^{-1})^2 = 36\text{ J}

At the bottom of the incline, the remaining mechanical energy is kinetic energy.

3

Apply the work-energy theorem to find the work done against friction (

W_f

).

W_f = E_p - E_k = 100\text{ J} - 36\text{ J} = 64\text{ J}

The non-conservative friction force reduces the total mechanical energy by doing work against motion.

Key Concept

Work-Energy Theorem and Conservation of Energy with Friction

Question 8Question →

A body of mass $2\text{ kg}$ is projected vertically upward from ground level with an initial speed of $30\text{ m s}^{-1}$ . During its entire flight, it experiences a constant resistive force due to air resistance of $5\text{ N}$ . Taking $g = 10\text{ m s}^{-2}$ , calculate the kinetic energy of the body in Joules when it returns to the ground level.

Show answer & explanation

Answer: 540

Answer

The kinetic energy of the body when it returns to ground level is

540\text{ J}

.

At launch, the body possesses an initial kinetic energy of

E_i = \frac{1}{2}(2)(30)^2 = 900\text{ J}

. During ascent, both gravity (

20\text{ N}

) and air resistance (

5\text{ N}

) retard the motion, giving a net downward force of

25\text{ N}

and a deceleration of

12.5\text{ m s}^{-2}

. The maximum height reached is

h = \frac{30^2}{2(12.5)} = 36\text{ m}

. Since the body travels up and down, total distance covered is

72\text{ m}

. Non-conservative work done against air resistance is

W = 5\text{ N} \times 72\text{ m} = 360\text{ J}

. Therefore, the remaining kinetic energy upon returning to the ground is

900\text{ J} - 360\text{ J} = 540\text{ J}

.

Step-by-Step Solution

1

Calculate initial kinetic energy of launch

E_i = 900\text{ J}

Kinetic energy is given by

\frac{1}{2}m u^2 = \frac{1}{2}(2\text{ kg})(30\text{ m s}^{-1})^2 = 900\text{ J}

.

2

Determine maximum height reached during ascent

h = 36\text{ m}

Net upward retarding force

F = mg + F_{\text{air}} = 20 + 5 = 25\text{ N}

, yielding a deceleration

a = 12.5\text{ m s}^{-2}

. Using

0 = u^2 - 2ah

,

h = \frac{900}{25} = 36\text{ m}

.

3

Calculate energy lost to air resistance over total trajectory

W_{\text{air}} = 360\text{ J}

Air resistance acts continuously over both ascent and descent (total distance

2h = 72\text{ m}

). Work dissipated

= F_{\text{air}} \times 2h = 5 \times 72 = 360\text{ J}

.

4

Subtract non-conservative work loss from initial mechanical energy

E_f = 540\text{ J}

By mechanical energy balance, final kinetic energy

E_f = E_i - W_{\text{air}} = 900 - 360 = 540\text{ J}

.

Key Concept

Work-Energy Theorem and Mechanical Energy Dissipation by Non-Conservative Forces

Question 9Question →

A spring with a force constant of $400\text{ N m}^{-1}$ is compressed by $0.1\text{ m}$ from its uncompressed length. When released, it launches a block of mass $0.16\text{ kg}$ along a smooth horizontal surface. What is the speed of the block immediately after leaving the spring?

5.0\text{ m s}^{-1}

2.5\text{ m s}^{-1}

12.5\text{ m s}^{-1}

25.0\text{ m s}^{-1}

Show answer & explanation

Answer: $5.0\text{ m s}^{-1}$

Answer

The speed of the block immediately after leaving the spring is

5.0\text{ m s}^{-1}

.

According to the principle of conservation of mechanical energy, the elastic potential energy stored in the compressed spring (

E_p = \frac{1}{2} k x^2

) is completely converted into kinetic energy (

E_k = \frac{1}{2} m v^2

) when the spring is released on a frictionless surface. Substituting

k = 400\text{ N m}^{-1}

,

x = 0.1\text{ m}

, and

m = 0.16\text{ kg}

yields

2.0\text{ J} = 0.08 v^2

, which solves to

v = 5.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the elastic potential energy stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 400 \times (0.1)^2 = 200 \times 0.01 = 2.0\text{ J}

Work done in compressing the spring is stored as elastic potential energy.

2

Apply conservation of mechanical energy.

E_k = E_p \implies \frac{1}{2} m v^2 = 2.0\text{ J}

Since the surface is smooth, all potential energy transforms into kinetic energy.

3

Solve for the velocity

v

of the block.

\frac{1}{2} \times 0.16 \times v^2 = 2.0 \implies 0.08 v^2 = 2.0 \implies v^2 = 25 \implies v = 5.0\text{ m s}^{-1}

Isolate

v

by dividing by

0.08

and taking the square root.

Key Concept

Conservation of Mechanical Energy (Elastic Potential Energy to Kinetic Energy)

Question 10Question →

A crate is pulled along a smooth horizontal floor by a constant force of $80\text{ N}$ applied at an angle of $60^\circ$ to the horizontal. If the crate moves through a displacement of $15\text{ m}$ along the floor, what is the work done by the applied force?

600\text{ J}

1039\text{ J}

1200\text{ J}

300\text{ J}

Show answer & explanation

Answer: $600\text{ J}$

Answer

600\text{ J}

The work done by a constant force applied at an angle

\theta

to the direction of motion is given by

W = F s \cos\theta

. Substituting

F = 80\text{ N}

,

s = 15\text{ m}

, and

\cos(60^\circ) = 0.5

gives

W = 80 \times 15 \times 0.5 = 600\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Force

F = 80\text{ N}

, displacement

s = 15\text{ m}

, angle

\theta = 60^\circ

Extract parameters required for the work formula.

2

Apply the work formula for a force at an angle

W = F s \cos\theta

Only the component of force parallel to the displacement does work on the object.

3

Substitute the values and compute the result

W = 80 \times 15 \times \cos(60^\circ) = 1200 \times 0.5 = 600\text{ J}

Calculates the effective work done by the applied force.

Key Concept

Work Done by a Constant Force at an Angle

Estimated Time:1m 0s

Question 11Question →

An electric motor with an efficiency of $80\%$ is used to pull a $100\text{ kg}$ object up a smooth incline inclined at $30^\circ$ to the horizontal at a constant speed of $2\text{ m s}^{-1}$ . Taking acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the electrical power input to the motor, in watts?

Show answer & explanation

Answer: 1250

Answer

1250 W

The force needed to move the mass up the smooth incline at constant speed is the parallel component of weight,

F = mg \sin(30^\circ) = 500\text{ N}

. The useful power output is

P_{\text{out}} = Fv = 500 \times 2 = 1000\text{ W}

. Dividing by the efficiency of

0.80

yields the electrical power input

P_{\text{in}} = 1250\text{ W}

.

Step-by-Step Solution

1

Determine the force required along the inclined plane.

F = mg \sin(30^\circ) = 100\text{ kg} \times 10\text{ m s}^{-2} \times 0.5 = 500\text{ N}

At constant velocity, the applied force balances the component of weight parallel to the incline.

2

Calculate the useful power output delivered by the motor.

P_{\text{out}} = F \times v = 500\text{ N} \times 2\text{ m s}^{-1} = 1000\text{ W}

Mechanical power output is the product of pulling force and constant speed.

3

Calculate the total electrical power input required.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{1000\text{ W}}{0.80} = 1250\text{ W}

Efficiency is defined as the ratio of useful power output to total power input.

Key Concept

Mechanical power on inclined planes and system efficiency

Question 12Question →

A water pump raises $600\text{ kg}$ of water through a vertical height of $20\text{ m}$ in $50\text{ s}$ . If the efficiency of the pump is $80\%$ , what is the electrical power input required to operate the pump? (Take $g = 10\text{ m s}^{-2}$ )

3.0\text{ kW}

2.4\text{ kW}

1.92\text{ kW}

120.0\text{ kW}

Show answer & explanation

Answer: $3.0\text{ kW}$

Answer

3.0\text{ kW}

The correct answer is

3.0\text{ kW}

. Raising

600\text{ kg}

of water by

20\text{ m}

requires

120,000\text{ J}

of gravitational potential energy. Doing this in

50\text{ s}

requires an output power of

2,400\text{ W}

(

2.4\text{ kW}

). Since the pump operates at

80\%

efficiency, the input power must be

\frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

.

Step-by-Step Solution

1

Calculate the useful work output needed to elevate the water

W_{out} = mgh = 600\text{ kg} \times 10\text{ m s}^{-2} \times 20\text{ m} = 120,000\text{ J}

Work done against gravity equals the gravitational potential energy gained

2

Calculate the useful output power of the pump

P_{out} = \frac{W_{out}}{t} = \frac{120,000\text{ J}}{50\text{ s}} = 2,400\text{ W} = 2.4\text{ kW}

Power is defined as the rate at which work is performed

3

Calculate the required electrical power input using efficiency

P_{in} = \frac{P_{out}}{\text{Efficiency}} = \frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

Efficiency is the ratio of useful output power to total input power

Key Concept

Work done against gravity, power, and mechanical/electrical efficiency

Estimated Time:1m 30s

Question 13Question →

A constant horizontal force acts on a body of mass $10\text{ kg}$ , accelerating it from rest to a speed of $12\text{ m s}^{-1}$ in a time of $4\text{ s}$ along a smooth horizontal surface. What is the average power delivered by the force during this time interval?

Show answer & explanation

Answer: 180

Answer

The average power delivered by the force during the 4-second interval is

180\text{ W}

.

By the work-energy theorem, the total work done by the constant force equals the gain in kinetic energy:

W = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times 144 = 720\text{ J}

. The average power is the rate at which work is performed over time:

P = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Step-by-Step Solution

1

Calculate the final kinetic energy acquired by the body.

E_k = \frac{1}{2} m v^2 = \frac{1}{2} (10\text{ kg})(12\text{ m s}^{-1})^2 = 720\text{ J}

.

Since the body starts from rest on a smooth surface, all work done by the net force goes into increasing its kinetic energy.

2

Divide the total work done by the elapsed time to find the average power.

P_{\text{avg}} = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Average power is defined as the rate of doing work over a given time interval.

Key Concept

Work-Energy Theorem and Average Power

Question 14Question →

An electric crane lifts a load of $250\text{ kg}$ vertically upwards through a height of $12\text{ m}$ in $10\text{ s}$ at a constant speed. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the useful output power of the crane in watts?

Show answer & explanation

Answer: 3000

Answer

The useful output power of the crane is

3000\text{ W}

.

The work done in lifting the load vertically is equal to the gravitational potential energy gained,

W = mgh = 250 \times 10 \times 12 = 30,000\text{ J}

. Power is the rate of doing work, so

P = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}

.

Step-by-Step Solution

1

Identify the given values

Mass

m = 250\text{ kg}

, height

h = 12\text{ m}

, time

t = 10\text{ s}

, acceleration due to gravity

g = 10\text{ m s}^{-2}

.

Extract values needed for work and power calculations.

2

Calculate the work done in lifting the load

W = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}

.

The work done against gravity equals the gain in gravitational potential energy.

3

Calculate the power output

P = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}

.

Power is defined as the rate at which work is done (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

Question 15Question →

A constant horizontal force of $25\text{ N}$ is applied to push a box across a smooth horizontal surface through a displacement of $8\text{ m}$ in the direction of the force. What is the total work done on the box?

200\text{ J}

40\text{ J}

3.125\text{ J}

33\text{ J}

Show answer & explanation

Answer: $200\text{ J}$

Answer

The work done on the box is

200\text{ J}

.

Work done is defined as the product of force and distance moved in the direction of the force (

W = F \times d

). Substituting the given values

F = 25\text{ N}

and

d = 8\text{ m}

gives

W = 200\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Force

F = 25\text{ N}

, displacement

d = 8\text{ m}

, angle

\theta = 0^\circ

Work depends on force and displacement along the line of action

2

Apply the work formula

W = F \cdot d \cos(\theta)

W = 25\text{ N} \times 8\text{ m} \times \cos(0^\circ) = 200\text{ J}

Since force and displacement are in the same direction,

\cos(0^\circ) = 1

Key Concept

Work Done by a Constant Force

Question 16Question →

A toy car of mass $0.5\text{ kg}$ moves along a smooth horizontal surface at a constant speed of $4\text{ m s}^{-1}$ . What is the kinetic energy of the toy car?

1\text{ J}

4\text{ J}

8\text{ J}

16\text{ J}

Show answer & explanation

Answer: $4\text{ J}$

Answer

The kinetic energy of the toy car is

4\text{ J}

.

Substituting the given values into the formula

E_k = \frac{1}{2} m v^2

gives

E_k = \frac{1}{2} \times 0.5 \times 4^2 = 4\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 0.5\text{ kg}

and speed

v = 4\text{ m s}^{-1}

.

These are the mandatory inputs needed to determine kinetic energy.

2

Apply the kinetic energy formula

E_k = \frac{1}{2} m v^2

Kinetic energy of a translational body is defined as half the product of its mass and the square of its speed.

3

Substitute the values and compute the result

E_k = \frac{1}{2} \times 0.5\text{ kg} \times (4\text{ m s}^{-1})^2 = 0.25 \times 16 = 4\text{ J}

.

Evaluating the mathematical operation yields the energy in Joules.

Key Concept

Kinetic Energy

Estimated Time:45s

Question 17Question →

A conveyor system pulls a $40\text{ kg}$ crate at a constant speed of $3\text{ m s}^{-1}$ up a rough inclined ramp. The ramp rises $3\text{ m}$ for every $5\text{ m}$ measured along its slope (giving $\sin\theta = 0.6$ and $\cos\theta = 0.8$ ). If the coefficient of kinetic friction between the crate and the ramp is $0.25$ and $g = 10\text{ m s}^{-2}$ , what is the power output of the conveyor system in watts?

Show answer & explanation

Answer: 960

Answer

The power output required by the conveyor system is

960\text{ W}

.

To pull the crate up the ramp at constant speed, the conveyor force must overcome both the parallel gravitational component (

240\text{ N}

) and friction (

80\text{ N}

), making the total force

320\text{ N}

. Multiplying this force by the constant speed of

3\text{ m s}^{-1}

gives a total power output of

960\text{ W}

.

Step-by-Step Solution

1

Calculate the component of weight parallel to the inclined plane

F_g = mg \sin\theta = 40 \times 10 \times 0.6 = 240\text{ N}

Gravity pulls the object back down along the slope with force

mg \sin\theta

.

2

Calculate the normal reaction force perpendicular to the plane

N = mg \cos\theta = 40 \times 10 \times 0.8 = 320\text{ N}

The normal force balances the perpendicular weight component.

3

Determine the magnitude of kinetic friction force

f_k = \mu N = 0.25 \times 320 = 80\text{ N}

Friction opposes motion up the slope and depends on the normal force.

4

Calculate the total pulling force needed for zero net acceleration

F = F_g + f_k = 240 + 80 = 320\text{ N}

At constant velocity, net force along the incline must equal zero, so

F = mg\sin\theta + f_k

.

5

Calculate the power output of the conveyor

P = F \times v = 320\text{ N} \times 3\text{ m s}^{-1} = 960\text{ W}

Power developed by a constant force moving an object at velocity

v

is given by

P = Fv

.

Key Concept

Work done against gravity and friction, and rate of doing work (Power

P = Fv

)

Question 18Question →

A car of mass $1200\text{ kg}$ ascends a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$ , at a steady speed of $15\text{ m s}^{-1}$ . If the total frictional resistance to motion is $400\text{ N}$ , what is the useful mechanical power output of the engine in kilowatts? (Take $g = 10\text{ m s}^{-2}$ )

Show answer & explanation

Answer: 24

Answer

24 kW

To maintain a constant ascending speed, the engine must supply a force equal to the sum of the component of weight parallel to the incline (

mg\sin\theta = 1200\text{ N}

) and the frictional force (

400\text{ N}

), resulting in a total force of

1600\text{ N}

. Multiplying this total force by the constant speed of

15\text{ m s}^{-1}

yields a power output of

24000\text{ W}

, which corresponds to

24\text{ kW}

.

Step-by-Step Solution

1

Calculate the gravitational force component acting down the slope

1200 N

The component of the car's weight parallel to the incline opposes upward motion:

F_g = m g \sin\theta = 1200 \text{ kg} \times 10 \text{ m s}^{-2} \times 0.1 = 1200 \text{ N}

.

2

Determine the total tractive force required from the car engine

1600 N

Because the velocity is constant, the net force is zero; hence, the engine force must balance both the gravitational slope component and the frictional resistance:

F_{\text{engine}} = F_g + F_{\text{friction}} = 1200 \text{ N} + 400 \text{ N} = 1600 \text{ N}

.

3

Calculate the mechanical power delivered by the engine

24 kW

Power is the product of tractive force and constant speed:

P = F_{\text{engine}} \times v = 1600 \text{ N} \times 15 \text{ m s}^{-1} = 24000 \text{ W} = 24 \text{ kW}

.

Key Concept

Power required to maintain motion against opposing forces on an inclined plane

Question 19Question →

A porter carries a suitcase of mass $5\text{ kg}$ along a horizontal platform for a distance of $10\text{ m}$ at a constant speed, and then lifts it vertically upward through a height of $2\text{ m}$ onto a shelf. Taking $g = 10\text{ m s}^{-2}$ , what is the total work done by the porter on the suitcase?

100\text{ J}

600\text{ J}

500\text{ J}

10\text{ J}

Show answer & explanation

Answer: $100\text{ J}$

Answer

100\text{ J}

During horizontal motion at constant speed, the upward force exerted by the porter is perpendicular to the horizontal displacement, so no work is performed horizontally (

W_{\text{horizontal}} = F s \cos 90^\circ = 0\text{ J}

). During vertical lifting, the force acts in the direction of displacement, yielding

W_{\text{vertical}} = mgh = 5\text{ kg} \times 10\text{ m s}^{-2} \times 2\text{ m} = 100\text{ J}

. The total work done is therefore

100\text{ J}

.

Step-by-Step Solution

1

Calculate the work done during horizontal motion

W_{\text{horizontal}} = 0\text{ J}

The supporting force acts vertically upward at an angle of

90^\circ

to the horizontal displacement, giving

W = F s \cos 90^\circ = 0\text{ J}

.

2

Calculate the work done in lifting the suitcase vertically

W_{\text{vertical}} = mgh = 5 \times 10 \times 2 = 100\text{ J}

Work done against gravity equals the increase in gravitational potential energy.

3

Sum the work done in both stages

W_{\text{total}} = 0 + 100 = 100\text{ J}

Total work done is the scalar sum of work performed along each segment of motion.

Key Concept

Work done by a constant force depends on the direction of displacement (

W = F s \cos \theta

). Perpendicular forces do no work.

Question 20Question →

An object of mass $4\text{ kg}$ is released from rest at a height of $5\text{ m}$ above the ground. Neglecting air resistance and taking $g = 10\text{ m s}^{-2}$ , what is its kinetic energy just before striking the ground?

200\text{ J}

100\text{ J}

40\text{ J}

20\text{ J}

Show answer & explanation

Answer: $200\text{ J}$

Answer

The kinetic energy of the object just before striking the ground is

200\text{ J}

.

The total mechanical energy is conserved during free fall. The initial gravitational potential energy

E_p = mgh = 4 \times 10 \times 5 = 200\text{ J}

is completely converted into kinetic energy

E_k

just before impact, making

200\text{ J}

the correct answer.

Step-by-Step Solution

1

Calculate the initial potential energy at maximum height

E_p = mgh = 4\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 200\text{ J}

At the top of the fall, all mechanical energy is stored as gravitational potential energy.

2

Apply the principle of conservation of mechanical energy

E_k = E_p = 200\text{ J}

In the absence of resistive forces such as air drag, potential energy lost converts completely into kinetic energy gained.

Key Concept

Conservation of Mechanical Energy

Estimated Time:45s

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