If a machine does 300 J of work in 5 seconds, what is its power output?

Correct Option: A. Solution: Power is calculated as $P = \frac{W}{t} = \frac{300}{5} = 60 \text{ W}$.

A windmill with blade area $A$ is exposed to wind flowing at velocity $v$. If the density of air is $\rho$, what is the expression for the mass of air passing through the windmill in time $t$?

Correct Option: B. Solution: The volume of air passing through the windmill in time $t$ is $Avt$. The mass of the air is then the volume times the density, $\rho Avt$.

Work, Energy and Power

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the work-energy theorem and its implications.
Define work and calculate it for various forces.
Explain kinetic energy and its relationship with work.
Analyze work done by variable forces and apply the work-energy theorem.
Describe potential energy and its conservation in mechanical systems.
Calculate the potential energy of springs and understand their behavior.
Define power and calculate it in different contexts.
Understand the principles of collisions and their types (elastic and inelastic).
Apply conservation laws to solve problems involving collisions.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Work, Energy, and Power

5.1 Introduction

5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem

The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body.

5.3 Work

Work done by a force is defined as the product of the force and the displacement in the direction of the force.

5.4 Kinetic Energy

Kinetic energy (K) is given by the formula:
- K = 0.5 * m * v²

5.5 Work Done by a Variable Force

The work done by a variable force can be calculated using the area under the force vs. distance graph.

5.6 The Work-Energy Theorem for a Variable Force

The work done by a variable force is equal to the change in kinetic energy.

5.7 The Concept of Potential Energy

Potential energy (V) is energy stored due to an object's position or configuration.

5.8 The Conservation of Mechanical Energy

The total mechanical energy of a body remains constant if only conservative forces act on it.

5.9 The Potential Energy of a Spring

The potential energy stored in a spring is given by:
- V(x) = 0.5 * k * x²

5.10 Power

Power (P) is defined as the rate at which work is done:
- P = W/t

5.11 Collisions

In elastic collisions, both momentum and kinetic energy are conserved.

Summary

Work is the energy transferred by a force.
Kinetic energy is energy of motion.
Potential energy is stored energy based on position.
Power measures how quickly work is done.

Points to Ponder

Consider the implications of energy conservation in real-world scenarios.

Exercises

Calculate work done in various scenarios involving lifting and friction.
Analyze potential energy functions and their implications.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Work and Energy: Students often confuse work done with energy transferred. Remember that work is the energy transferred by a force over a distance.
Forgetting the Direction of Forces: When calculating work, ensure that the direction of the force and the displacement are considered. Work can be negative if the force opposes the displacement.
Confusing Elastic and Inelastic Collisions: Students may not clearly differentiate between elastic and inelastic collisions. Remember that in elastic collisions, kinetic energy is conserved, while inelastic collisions do not conserve kinetic energy.
Ignoring Units: Always check that your units are consistent, especially when calculating power and energy. For example, power is measured in watts (W), which is equivalent to joules per second (J/s).

Tips for Success

Review the Work-Energy Theorem: Understand that the work done by the net force on an object is equal to the change in its kinetic energy. This can help in solving problems related to motion and forces.
Practice Collision Problems: Work through various problems involving elastic and inelastic collisions to solidify your understanding of momentum conservation.
Use Diagrams: When dealing with forces and motion, sketching diagrams can help visualize the problem and clarify the relationships between different quantities.
Memorize Key Formulas: Familiarize yourself with essential formulas such as the work-energy theorem, conservation of momentum, and the definitions of kinetic and potential energy.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

\sqrt{\frac{k}{m}} x_0

\sqrt{\frac{2k}{m}} x_0

\sqrt{\frac{k}{2m}} x_0

\sqrt{\frac{k}{m}} \frac{x_0}{2}

Correct Answer: A

Solution:

The total mechanical energy

E

is conserved. Initially, all energy is potential:

E = \frac{1}{2}kx_0^2

. At

x = 0

, all energy is kinetic:

\frac{1}{2}mv^2 = \frac{1}{2}kx_0^2

. Solving for

v

, we get

v = \sqrt{\frac{k}{m}} x_0

750 W

1500 W

3000 W

3750 W

Correct Answer: B

Solution:

The mass flow rate of air is

\rho A v = 1.2 \times 50 \times 10 = 600 \text{ kg/s}

. The kinetic energy per second (power) of the wind is

\frac{1}{2} \times 600 \times 10^2 = 30000 \text{ W}

. The electrical power output is 25% of this, which is

0.25 \times 30000 = 7500 \text{ W}

20 J

60 J

40 J

80 J

Correct Answer: B

Solution:

Work done

W

is calculated as

W = F \cdot d = 20 \times 3 = 60 \text{ J}

0.5 m/s

1.0 m/s

2.0 m/s

5.0 m/s

Correct Answer: B

Solution:

The power

P = Fv

, where

F = mg = 200 \times 9.8 = 1960 \text{ N}

. Thus,

v = \frac{P}{F} = \frac{1000}{1960} \approx 0.51 \text{ m/s}

. The closest option is 1.0 m/s.

60 W

1500 W

300 W

75 W

Correct Answer: A

Solution:

Power is calculated as

P = \frac{W}{t} = \frac{300}{5} = 60 \text{ W}

30 W

3 W

3000 W

300 W

Correct Answer: A

Solution:

Power

P

is defined as the rate at which work is done. It is given by

P = \frac{W}{t}

, where

W

is the work done and

t

is the time taken. Substituting

W = 300 \text{ J}

and

t = 10 \text{ s}

, we get

P = \frac{300}{10} = 30 \text{ W}

30°

45°

60°

90°

Correct Answer: D

Solution:

In a perfectly elastic collision between two identical masses, if one is initially at rest, they will move at right angles to each other after the collision. Thus, the angle is 90°.

x = 2 \text{ m}

x = 1 \text{ m}

x = 3 \text{ m}

x = 0.5 \text{ m}

Correct Answer: A

Solution:

The particle will turn back when its kinetic energy becomes zero, i.e., when all its energy is potential. Using

V(x) = \frac{1}{2}kx^2 = 1 \text{ J}

, we solve for

x

1 = \frac{1}{2} \times 0.5 \times x^2

, giving

x = 2 \text{ m}

Both masses have velocity

v/2

The initially moving mass comes to rest, and the initially stationary mass moves with velocity

v

Both masses have velocity

v

The initially moving mass continues with velocity

v

, and the initially stationary mass remains at rest

Correct Answer: B

Solution:

In an elastic collision between two identical masses where one is initially at rest, the moving mass transfers all its velocity to the stationary mass. Therefore, the initially moving mass comes to rest, and the initially stationary mass moves with velocity

v

30°

45°

53°

60°

Correct Answer: C

Solution:

For two equal masses undergoing a glancing elastic collision, they move at right angles to each other after the collision. Therefore,

\theta_1 + 37° = 90°

, giving

\theta_1 = 53°

It rises to the same initial height.

It rises to half the initial height.

It does not rise at all.

It rises to a height greater than the initial height.

Correct Answer: A

Solution:

In an elastic collision, the kinetic energy and momentum are conserved. Since both bobs have the same mass and the collision is elastic, the first bob will transfer all its energy to the second bob and will not rise. However, after the collision, the first bob will rise back to its initial height due to the conservation of energy.

10 J

20 J

50 J

100 J

Correct Answer: C

Solution:

The work done

W

is given by the formula

W = F \cdot d

. Substituting

F = 10 \text{ N}

and

d = 5 \text{ m}

, we get

W = 10 \times 5 = 50 \text{ J}

\sqrt{2gh}

\sqrt{gh}

\sqrt{4gh}

\sqrt{\frac{gh}{2}}

Correct Answer: A

Solution:

The potential energy at the release height is converted entirely into kinetic energy at the lowest point. Using the conservation of mechanical energy:

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

. Solving for

v

, we get

v = \sqrt{2gh}

500 W

490 W

50 W

100 W

Correct Answer: B

Solution:

Power

P

is given by

P = F \cdot v

, where

F

is the force and

v

is the velocity. Here,

F = mg = 10 \times 9.8 = 98 \text{ N}

. Thus,

P = 98 \times 5 = 490 \text{ W}

1.5 J

2.5 J

3.5 J

4.5 J

Correct Answer: B

Solution:

The total mechanical energy

E

is the sum of kinetic energy

K

and potential energy

V

. The kinetic energy

K = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (2)^2 = 2 \text{ J}

. The potential energy

V = \frac{1}{2}kx^2 = \frac{1}{2} \times 0.5 \times (1)^2 = 0.25 \text{ J}

. Thus,

E = K + V = 2 + 0.25 = 2.5 \text{ J}

27 km/h

30 km/h

25 km/h

20 km/h

Correct Answer: A

Solution:

Since the track is frictionless and no external forces act on the system, the speed of the trolley remains constant at 27 km/h.

x_{max} = \sqrt{\frac{2E}{k}}

x_{max} = \sqrt{\frac{E}{k}}

x_{max} = \frac{E}{k}

x_{max} = \frac{2E}{k}

Correct Answer: A

Solution:

The total mechanical energy

E

is the sum of kinetic and potential energy. At maximum displacement, kinetic energy is zero, so

E = \frac{1}{2}kx_{max}^2

. Solving for

x_{max}

gives

x_{max} = \sqrt{\frac{2E}{k}}

It remains the same

It doubles

It quadruples

It halves

Correct Answer: A

Solution:

The gravitational force depends only on the masses and the distance between them, not on the speed of the satellite. Therefore, the gravitational force remains the same.

0 degrees

30 degrees

45 degrees

90 degrees

Correct Answer: D

Solution:

In an elastic collision between two equal masses, the velocities after the collision are at right angles (90 degrees) to each other.

1 J

2 J

0.5 J

0.1 J

Correct Answer: C

Solution:

The potential energy stored in a spring is given by

V = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \text{ J}

The work done is positive.

The work done is negative.

The work done is zero.

The work done depends on the speed of the satellite.

Correct Answer: C

Solution:

In a circular orbit, the gravitational force is always perpendicular to the velocity of the satellite. Therefore, the work done by the gravitational force over one complete orbit is zero.

2 m

1.41 m

1 m

2.83 m

Correct Answer: B

Solution:

The total mechanical energy

E

is given by

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

. Solving for

x

, we have

2 = \frac{1}{2} \cdot 1 \cdot x^2 \Rightarrow x^2 = 4 \Rightarrow x = \sqrt{4} = 2

. Therefore, the maximum displacement is

x = \frac{2}{\sqrt{2}} = 1.41 \text{ m}

27 km/h

30 km/h

24 km/h

33 km/h

Correct Answer: A

Solution:

Since the track is frictionless and no external forces act on the system, the momentum of the system is conserved. The initial momentum is

(300 + 25) \times 27

. As the sand leaks out, the trolley's speed remains constant at 27 km/h because the leaking sand does not exert any horizontal force on the trolley.

1000 J

1000 kWh

1 kWh

10 kWh

Correct Answer: C

Solution:

Energy consumed is calculated as power multiplied by time. Therefore, 100 watts for 10 hours is 1000 watt-hours, which is 1 kWh.

Avt

\rho Avt

\frac{1}{2} \rho Avt

\rho A^2vt

Correct Answer: B

Solution:

The volume of air passing through the windmill in time

t

Avt

. The mass of the air is then the volume times the density,

\rho Avt

50 J

75 J

100 J

125 J

Correct Answer: A

Solution:

The work done is equal to the change in kinetic energy. The kinetic energy

K

is given by

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

. Substituting

v = ax^{3/2}

, we have

K = \frac{1}{2}m(a^2x^3)

. The work done is

\Delta K = \frac{1}{2}m(a^2(2)^3) - \frac{1}{2}m(a^2(0)^3) = \frac{1}{2} \times 0.5 \times (5^2 \times 8) = 50

100 W

50 W

125 W

75 W

Correct Answer: A

Solution:

Power

P

is calculated as

P = \frac{W}{t}

, where

W = 500 \text{ J}

and

t = 5 \text{ s}

. Thus,

P = \frac{500}{5} = 100 \text{ W}

The work done by it is path independent.

It always does positive work.

It does not do any work.

It only acts vertically.

Correct Answer: A

Solution:

A conservative force is one where the work done is path independent and depends only on the initial and final positions.

15 J

22.5 J

45 J

30 J

Correct Answer: B

Solution:

The kinetic energy is given by

K = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times 3^2 = 22.5 \text{ J}

5 W

50 W

500 W

5000 W

Correct Answer: B

Solution:

Power

P

is given by the formula

P = \frac{W}{t}

. Substituting

W = 500 \text{ J}

and

t = 10 \text{ s}

, we get

P = \frac{500}{10} = 50 \text{ W}

1 J

0.5 J

2 J

10 J

Correct Answer: B

Solution:

The potential energy

V

of a spring is given by

V = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 0.5 \text{ J}

-9 J

-4.5 J

0 J

9 J

Correct Answer: A

Solution:

The work done by a conservative force is given by the negative change in potential energy. Here,

V(x) = \frac{1}{2}kx^2

. The change in potential energy from

x = 0

x = 3 \text{ m}

\Delta V = \frac{1}{2} \times 2 \times (3^2 - 0^2) = 9 \text{ J}

. Thus, the work done is

-9 \text{ J}

6.26 m/s

4.43 m/s

5.56 m/s

6.32 m/s

Correct Answer: B

Solution:

Using conservation of energy, the potential energy at the top is converted to kinetic energy at the bottom.

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

. Solving for

v

v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2} = 4.43 \text{ m/s}

10 m/s

5 m/s

0 m/s

20 m/s

Correct Answer: A

Solution:

In a perfectly elastic collision between two identical masses, the moving ball transfers all its velocity to the stationary ball. Hence, the second ball moves with a speed of 10 m/s.

0.065 kg

0.130 kg

0.260 kg

0.520 kg

Correct Answer: A

Solution:

The work done against gravity each time is

W = mgh = 10 \times 9.8 \times 0.5 = 49

J. For 1000 lifts, the total work done is

49,000

J. With a conversion efficiency of 20%, the energy used from fat is

\frac{49,000}{0.2} = 245,000

J. The amount of fat used is

\frac{245,000}{3.8 \times 10^7} \approx 0.065

kg.

1 J

2 J

4 J

8 J

Correct Answer: A

Solution:

The potential energy stored in a spring is given by

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

. Substituting

k = 100 \text{ N/m}

and

x = 0.2 \text{ m}

, we get

V = \frac{1}{2} \times 100 \times (0.2)^2 = 2 \text{ J}

3 J

6 J

9 J

18 J

Correct Answer: C

Solution:

Kinetic energy

K

is given by

K = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 3^2 = 9 \text{ J}

49 J

98 J

490 J

980 J

Correct Answer: C

Solution:

The potential energy

V

is given by

V = mgh = 10 \times 9.8 \times 5 = 490 \text{ J}

It depends only on the initial and final positions.

It depends on the path taken.

It is always zero.

It is always positive.

Correct Answer: A

Solution:

A conservative force does work that is path-independent and depends only on the initial and final positions.

50 J

100 J

150 J

200 J

Correct Answer: B

Solution:

The work done by the net force is calculated by integrating the force over the displacement. Given

v = ax^{3/2}

, we find the force and integrate from

x = 0

x = 2 \text{ m}

to find the work done, which is 100 J.

\frac{32}{5}ma^2

\frac{64}{5}ma^2

\frac{128}{5}ma^2

\frac{256}{5}ma^2

Correct Answer: B

Solution:

The velocity is given as

v = ax^{3/2}

. The kinetic energy

K = \frac{1}{2}mv^2 = \frac{1}{2}m(a^2x^3)

. The work done by the net force is the change in kinetic energy from

x=0

x=4

. At

x=0

K_i = 0

. At

x=4

K_f = \frac{1}{2}m(a^2(4)^3) = 32ma^2

. Therefore, the work done

W = K_f - K_i = 32ma^2

37°

45°

53°

60°

Correct Answer: C

Solution:

In an elastic collision between two equal masses, the angle between their velocities after the collision is 90°. Using the conservation of momentum and kinetic energy, the angle for the first ball is

53°

5000 J

49000 J

4900 J

500 J

Correct Answer: B

Solution:

Work done against gravity is calculated as

W = mgh

. For each lift,

W = 10 \times 9.8 \times 0.5 = 49 \text{ J}

. For 1000 lifts, total work is

49 \times 1000 = 49000 \text{ J}

It is always conserved in any collision.

It is a scalar quantity.

It depends on the direction of motion.

It is measured in newtons.

Correct Answer: B

Solution:

Kinetic energy is a scalar quantity and is measured in joules.

5 J

45 J

30 J

60 J

Correct Answer: B

Solution:

Work done is calculated as

W = F \cdot d = 15 \times 3 = 45 \text{ J}

The gravitational force increases.

The satellite's potential energy decreases more than its kinetic energy.

The satellite's mass decreases.

The atmospheric drag reduces its velocity.

Correct Answer: B

Solution:

As the satellite loses energy, it moves to a lower orbit where its potential energy decreases. The decrease in potential energy is greater than the decrease in total energy, causing an increase in kinetic energy and thus speed.

14 m/s

10 m/s

20 m/s

28 m/s

Correct Answer: A

Solution:

Using conservation of mechanical energy, the potential energy at the top is converted to kinetic energy at the bottom.

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

. Solving for

v

v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 10} = 14 \text{ m/s}

\rho A v t

\frac{1}{2} \rho A v t

2 \rho A v t

\rho A v^2 t

Correct Answer: A

Solution:

The mass of the air passing through the area

A

in time

t

is given by the product of the volume of air and its density. The volume of air is

A v t

, and thus the mass is

\rho A v t

, where

\rho

is the density of air.

The speed of the trolley increases.

The speed of the trolley decreases.

The speed of the trolley remains constant.

The speed of the trolley fluctuates.

Correct Answer: C

Solution:

Since the track is frictionless and no external horizontal forces act on the system, the speed of the trolley remains constant despite the loss of mass.

Joule

Newton

Watt

Horsepower

Correct Answer: C

Solution:

The SI unit of power is the watt (W), which is equivalent to one joule per second.

80 J

60 J

40 J

100 J

Correct Answer: A

Solution:

Work done

W

is calculated as

W = F \cdot d

, where

F = 20 \text{ N}

and

d = 4 \text{ m}

. Thus,

W = 20 \times 4 = 80 \text{ J}

The gravitational force decreases as it moves closer.

The potential energy decreases, converting to kinetic energy.

The satellite's mass decreases.

The drag force increases its kinetic energy.

Correct Answer: B

Solution:

As the satellite loses energy due to drag, it moves closer to Earth, decreasing its potential energy. According to the conservation of mechanical energy, this decrease in potential energy results in an increase in kinetic energy, thus increasing its speed.

8 J

16 J

32 J

64 J

Correct Answer: C

Solution:

The kinetic energy

K

is given by the formula

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

. Substituting

m = 2 \text{ kg}

and

v = 4 \text{ m/s}

, we get

K = \frac{1}{2} \times 2 \times 4^2 = 16 \text{ J}

x = \pm 2 \text{ m}

x = \pm \sqrt{2} \text{ m}

x = \pm 1 \text{ m}

x = 0 \text{ m}

Correct Answer: A

Solution:

The total energy

E

of the particle is given by

E = K + V(x)

. When the kinetic energy

K = 0

, the potential energy

V(x)

equals the total energy. Therefore,

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

. Solving for

x

, we get

x = \pm \sqrt{\frac{4}{1}} = \pm 2 \text{ m}

200,000 J

400,000 J

100,000 J

300,000 J

Correct Answer: A

Solution:

The kinetic energy

K

is given by

K = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times (20)^2 = 200,000 \text{ J}

44000 W

36000 W

50000 W

22000 W

Correct Answer: A

Solution:

The power output is

P = F \cdot v

. The force

F = mg + 4000 = 1800 \times 9.8 + 4000 = 22000 \text{ N}

. Thus,

P = 22000 \times 2 = 44000 \text{ W}

Because gravitational potential energy decreases.

Because kinetic energy is converted into potential energy.

Because the mass of the satellite increases.

Because the satellite is gaining energy from the atmosphere.

Correct Answer: A

Solution:

As the satellite loses energy, it spirals closer to Earth, converting potential energy into kinetic energy, thus increasing its speed.

30°

45°

60°

90°

Correct Answer: D

Solution:

For two identical masses undergoing an elastic collision with one initially at rest, they move at right angles to each other after the collision, hence the angle is 90°.

20 W

30 W

40 W

50 W

Correct Answer: C

Solution:

Power

P

is given by the formula

P = \frac{W}{t}

, where

W

is the work done and

t

is the time taken. Substituting

W = 600 \text{ J}

and

t = 20 \text{ s}

, we get

P = \frac{600}{20} = 30 \text{ W}

5.4 m/s

5.2 m/s

5.0 m/s

4.8 m/s

Correct Answer: A

Solution:

The initial potential energy is

mgh

. After losing 5% to air resistance, 95% of this energy is converted to kinetic energy at the lowest point. Solving for speed,

v = \sqrt{2gh \times 0.95}

, gives

v \approx 5.4 \text{ m/s}

0.5 J

1 J

2 J

4 J

Correct Answer: B

Solution:

The potential energy stored in a spring is given by

V(x) = \frac{1}{2} k x^2

. Substituting

k = 0.5 \text{ N m}^{-1}

and

x = 2 \text{ m}

, we find

V(x) = 1 \text{ J}

2500 J

5000 J

1000 J

2000 J

Correct Answer: B

Solution:

The kinetic energy

K

is given by

K = \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times (10)^2 = 2500 \text{ J}

20 J

40 J

10 J

5 J

Correct Answer: B

Solution:

The kinetic energy

K

is given by the formula

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

. Substituting

m = 10 \text{ kg}

and

v = 2 \text{ m/s}

, we get

K = \frac{1}{2} \times 10 \times 2^2 = 20 \times 2 = 40 \text{ J}

20 J

40 J

50 J

80 J

Correct Answer: B

Solution:

The work done by the force is equal to the change in kinetic energy. Work done

W = F \times d = 10 \times 4 = 40

Solution:

The potential energy function for simple harmonic motion is indeed

Solution:

The work done by friction over a closed path is not zero, and no potential energy can be associated with friction.

Correct Answer: True

Solution:

The scalar product

A \cdot B = AB \cos \theta

can be negative if the angle

\theta

between the vectors is greater than

90^\circ

Correct Answer: True

Solution:

The principle of conservation of mechanical energy states that the total mechanical energy remains constant if only conservative forces are acting.

Correct Answer: False

Solution:

The total mechanical energy of a system is conserved only if the forces acting on it are conservative.

Correct Answer: True

Solution:

A conservative force is defined such that the work done by it on an object is path-independent and depends only on the initial and final positions.

Correct Answer: True

Solution:

This is the standard formula for the potential energy of a particle executing simple harmonic motion, where

Solution:

The potential energy stored in a spring is calculated using the formula

Solution:

The gravitational potential energy is determined with respect to a reference point, often chosen as the ground level.

Correct Answer: True

Solution:

The potential energy function for a particle in simple harmonic motion is indeed given by

Solution:

The gravitational potential energy is calculated as

V(x) = mgx

, where

m

is the mass,

g

is the acceleration due to gravity, and

x

is the height.

Correct Answer: True

Solution:

A conservative force is defined such that the work done by it depends only on the initial and final positions, not on the path taken.

Correct Answer: False

Solution:

In a conservative force field, the work done by the force on an object is path-independent; it depends only on the initial and final positions.

Correct Answer: True

Solution:

The potential energy stored in a spring is described by the formula

Solution:

In an elastic collision, the total kinetic energy is conserved only before and after the collision, not during the collision process itself.

Correct Answer: False

Solution:

The rate of change of total momentum of a many-particle system is proportional to the external force on the system, not the sum of the internal forces.

Work, Energy and Power

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Work, Energy, and Power

5.1 Introduction

5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem

5.3 Work

5.4 Kinetic Energy

5.5 Work Done by a Variable Force

5.6 The Work-Energy Theorem for a Variable Force

5.7 The Concept of Potential Energy

5.8 The Conservation of Mechanical Energy

5.9 The Potential Energy of a Spring

5.10 Power

5.11 Collisions

Summary

Points to Ponder

Exercises

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Multiple Choice Questions

Correct Answer: A

Q2.A windmill converts 25% of the wind's kinetic energy into electrical energy. If the wind flows at a velocity of 10 m/s through an area of 50 m², and the density of air is 1.2 kg/m³, what is the electrical power output of the windmill?

Correct Answer: B

Q3.If a force of 20 N is applied to move a box 3 meters along a straight path, what is the work done by the force?

Correct Answer: B

Q4.A 1000 W motor is used to lift a 200 kg mass vertically upward at a constant speed. What is the maximum speed at which the motor can lift the mass?

Correct Answer: B

Q5.If a machine does 300 J of work in 5 seconds, what is its power output?

Correct Answer: A

Q6.A machine does 300 J of work in 10 seconds. What is its power output?

Correct Answer: A

Q7.In a perfectly elastic collision between two identical billiard balls, one of which is initially at rest, what is the angle between their velocities after the collision?

Correct Answer: D

Correct Answer: A

Q9.Two identical masses collide elastically in one dimension. If one mass is initially at rest and the other has a velocity vvv, what are their velocities after the collision?

Correct Answer: B

Q10.During an elastic collision between two equal masses, one of which is initially at rest, the first mass moves at an angle of θ1\theta_1θ1​ with respect to its original direction after the collision. What is the value of θ1\theta_1θ1​?

Correct Answer: C

Q11.A pendulum bob is released from a height where it makes an angle of 30° with the vertical. It collides elastically with another bob of the same mass at rest on a table. Assuming no energy is lost in air resistance, to what height does the first bob rise after the collision?

Correct Answer: A

Q12.If a force of 10 N is applied to move a box 5 meters along a straight path, what is the work done by the force?

Correct Answer: C

Q13.A pendulum bob of mass mmm is released from a height hhh above its lowest point. Assuming no air resistance, what is the speed of the bob at its lowest point?

Correct Answer: A

Q14.A 10 kg mass is lifted vertically with a constant speed of 5 m/s. If the gravitational force is the only force acting on the mass, what is the power required to maintain this motion?

Correct Answer: B

Correct Answer: B

Q16.A 300 kg trolley carrying a 25 kg sandbag moves at 27 km/h on a frictionless track. If sand leaks at 0.05 kg/s, what is the speed of the trolley after the sandbag is empty?

Correct Answer: A

Q17.A particle of mass mmm is moving in a potential field described by V(x)=12kx2V(x) = \frac{1}{2}kx^2V(x)=21​kx2. If the total mechanical energy of the particle is EEE, what is the maximum displacement xmaxx_{max}xmax​ from the equilibrium position?

Correct Answer: A

Q18.A satellite is orbiting Earth in a circular orbit. If the satellite's speed is doubled, what happens to the gravitational force acting on it?

Correct Answer: A

Q19.In an elastic collision between two equal masses where one mass is initially at rest, what is the angle between their velocities after the collision?

Correct Answer: D

Q20.A spring with a force constant of k=200 N/mk = 200 \text{ N/m}k=200 N/m is compressed by 0.1 m0.1 \text{ m}0.1 m. What is the potential energy stored in the spring?

Correct Answer: C

Q21.A satellite orbits Earth in a circular path with a constant speed. Which of the following statements is true about the work done by the gravitational force on the satellite?

Correct Answer: C

Correct Answer: B

Q23.A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. Sand leaks out at a rate of 0.05 kg/s. What is the speed of the trolley after the entire sandbag is empty?

Correct Answer: A

Q24.A 100 watt bulb is used for 10 hours. How much energy does it consume?

Correct Answer: C

Q25.A windmill with blade area AAA is exposed to wind flowing at velocity vvv. If the density of air is ρ\rhoρ, what is the expression for the mass of air passing through the windmill in time ttt?

Correct Answer: B

Q26.A body of mass 0.5 kg travels in a straight line with velocity given by v=ax3/2v = ax^{3/2}v=ax3/2, where a=5extm−1/2exts−1a = 5 ext{ m}^{-1/2} ext{ s}^{-1}a=5extm−1/2exts−1. What is the work done by the net force during its displacement from x=0x = 0x=0 to x=2x = 2x=2 m?

Correct Answer: A

Q27.A machine does 500 J of work in 5 seconds. What is its power output?

Correct Answer: A

Q28.A force is considered conservative if:

Correct Answer: A

Q29.A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

Correct Answer: B

Q30.What is the power output of a machine that does 500 J of work in 10 seconds?