Work, Energy, and Simple Machines Class 9

20 Questions | 70 Marks

Define work done by a constant force. Give the SI unit of work and define it. 3marks

A force of 7 N acting on a body produces an extension of 10 cm in a spring. How much work is done in stretching the spring by 15 cm? 3marks

Explain the conditions under which work done by a force can be zero. Give an example for each condition. 3marks

A body is moving on a rough horizontal surface. The work done by the force of friction is:(a) Positive(b) Negative(c) Zero(d) Undefined 1mark

State the work-energy theorem. Explain how it relates work done to the energy of an object. 2marks

A body of mass 10 kg is moving with a velocity of 20 m/s. If the mass of the body is doubled and its velocity is halved, what is the effect on its kinetic energy? 3marks

Calculate the kinetic energy of a car of mass 1500 kg moving at a speed of 60 km/h. 3marks

An object of mass 20 kg is lifted to a height of 5 m. Calculate its potential energy. If the object is let to fall, what will be its kinetic energy just before hitting the ground? (Assume g = 9.8 m s–2). 5marks

Two persons, each of mass 50 kg, climb up a rope by which they are attached to an electric motor. The first person climbs up the rope at a speed of 0.5 m/s and the second person climbs up at a speed of 2 m/s. What is the power dissipated by the motor in the two cases? 3marks

Define power. What is the SI unit of power? 2marks

Explain the concept of mechanical energy and the principle of conservation of mechanical energy. 3marks

A ball of mass 100 g moving at 10 m/s strikes a wall and rebounds with the same speed. Calculate the impulse experienced by the ball. 3marks

A boy of mass 50 kg climbs up a ladder of height 10 m in 20 seconds. What is the power generated by the boy? (Assume g = 10 m s–2). 3marks

Explain with the help of a diagram, how a fixed pulley makes work easier. What is its mechanical advantage? 3marks

A person uses an inclined plane of length 5 m to raise an object of mass 100 kg to a height of 2 m. Calculate the mechanical advantage of the inclined plane. If the person exerts a force of 400 N, is the inclined plane frictionless? (Assume g = 10 m s–2). 5marks

Explain the working of a lever with a suitable diagram and its principle. 3marks

What is a lever? Give two examples of levers and classify them. 3marks

A 50 kg man climbs a staircase of 50 steps in 25 seconds. If each step is 20 cm high, calculate the power of the man. (Assume g = 10 m s–2). 3marks

A cricket ball of mass 200 g moving at 20 m/s is brought to rest by a fielder in 0.05 seconds. Calculate the impulse and the average force applied by the fielder. 5marks

A family of four with a total mass of 200 kg is on vacation in a car of mass 1000 kg. They are travelling at a speed of 72 km/h. Suddenly, the driver applies the brakes and the car comes to a stop in 5 seconds. Calculate the kinetic energy of the car just before the brakes were applied and the average force exerted by the brakes. 5marks

Answers

Work done by a constant force is defined as the product of the force and the displacement of the object in the direction of the force.SI Unit of Work: Joule (J).1 Joule of work is done when a constant force of 1 Newton displaces an object by 1 metre in its own direction.

Extension in the spring,

x_{1} = 10

= 0.10

m. Force

F_{1} = 7

N.Work done

W = \frac{1}{2} F_{1} x_{1} = \frac{1}{2} \times 7 N \times 0.10 m = 0.35

J.This work done is stored as potential energy in the spring.Potential energy per unit extension

= \frac{W}{x_{1}} = \frac{0.35 J}{0.10 m} = 3.5

J/m.New extension,

x_{2} = 15

= 0.15

m.Work done in stretching the spring by 15 cm

= \frac{1}{2} F_{1} x_{1} \times \frac{x_{2}}{x_{1}} \times 2

(incorrect logic in NCERT for calculating force from extension)Correct approach: Spring constant

k = \frac{F_{1}}{x_{1}} = \frac{7 N}{0.10 m} = 70

N/m.Work done in stretching by

x_{2} = 15

= 0.15

m is

W_{2} = \frac{1}{2} k x_{2}^{2} = \frac{1}{2} \times 70 N/m \times (0.15 m)^{2} = 35 \times 0.0225 J = 0.7875

Work done is zero if:(a) Force is zero (

F = 0

): If no force is applied on an object, no work is done, even if it moves. For example, if an object moves due to inertia.(b) Displacement is zero (

s = 0

): If an object does not move, no work is done, no matter how large the force applied. Example: Pushing a wall.(c) Force and displacement are perpendicular to each other: If the force acting on an object is perpendicular to its displacement, the work done is zero. Example: A coolie carrying a load on his head walks on a horizontal platform. The force applied by the coolie is upwards, while the displacement is horizontal.

(b) Negative. Friction always opposes motion, so the work done by friction is always negative.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically,

W = Δ K = K_{f} - K_{i}

, where

K_{f}

is the final kinetic energy and

K_{i}

is the initial kinetic energy. This means that if positive work is done on an object, its kinetic energy increases, and if negative work is done, its kinetic energy decreases.

Initial kinetic energy,

K_{i} = \frac{1}{2} m v^{2}

.Mass

m = 10

kg, velocity

v = 20

m/s.

K_{i} = \frac{1}{2} \times 10 kg \times (20 m/s)^{2} = 5 \times 400 J = 2000

J.New mass

m^{'} = 2 m = 2 \times 10 kg = 20

kg.New velocity

v^{'} = \frac{v}{2} = \frac{20 m/s}{2} = 10

m/s.New kinetic energy,

K_{f} = \frac{1}{2} m^{'} (v^{'})^{2} = \frac{1}{2} \times 20 kg \times (10 m/s)^{2} = 10 \times 100 J = 1000

J.The new kinetic energy is half of the original kinetic energy (

1000

J vs

2000

J).

Mass of the car,

m = 1500

kg.Speed of the car,

v = 60

km/h.Convert speed to m/s:

v = 60 \times \frac{1000 m}{3600 s} = 60 \times \frac{5}{18} m/s = \frac{300}{18} m/s = \frac{50}{3}

m/s.Kinetic energy,

K = \frac{1}{2} m v^{2} = \frac{1}{2} \times 1500 kg \times {(\frac{50}{3} m/s)}^{2}

K = \frac{1}{2} \times 1500 \times \frac{2500}{9} J = 750 \times \frac{2500}{9} J = \frac{1875000}{9} J \approx 208333.33

Mass of the object,

m = 20

kg.Height,

h = 5

m.Acceleration due to gravity,

g = 9.8

m/s

^{2}

.Potential energy,

U = m g h = 20 kg \times 9.8 {m/s}^{2} \times 5 m = 980

J.When the object is let to fall, its potential energy is converted into kinetic energy. Just before hitting the ground, its potential energy is zero (assuming ground level is zero potential energy).By conservation of mechanical energy,

U_{i n i t i a l} + K_{i n i t i a l} = U_{f i n a l} + K_{f i n a l}

980

+ 0

= 0

+ K_{f i n a l}

.So, the kinetic energy just before hitting the ground is

K_{f i n a l} = 980

Mass of each person,

m = 50

kg.Let the height of the rope they climb be

h

.Case 1: Speed

v_{1} = 0.5

m/s.Work done per second by the motor = Power

= m g v_{1} = 50 kg \times 10 {m/s}^{2} \times 0.5 m/s = 250

W.Case 2: Speed

v_{2} = 2

m/s.Work done per second by the motor = Power

= m g v_{2} = 50 kg \times 10 {m/s}^{2} \times 2 m/s = 1000

Power is defined as the rate at which work is done. It is the work done per unit time.The SI unit of power is the Watt (W).

Mechanical energy is the sum of kinetic energy and potential energy of an object.

E = K + U

.The principle of conservation of mechanical energy states that if the only forces acting on an object are conservative forces (like gravity), then the total mechanical energy of the object remains constant. That is,

K + U =

constant. Energy can be transformed between kinetic and potential forms, but their sum stays the same.

Mass of the ball,

m = 100

= 0.1

kg.Initial velocity,

u = 10

m/s.Final velocity,

v = - 10

m/s (since it rebounds with the same speed in the opposite direction).Impulse

= Δ p = m v - m u = m (v - u)

Impulse

= 0.1 kg \times (- 10 m/s - 10 m/s) = 0.1 kg \times (- 20 m/s) = - 2

kg m/s.The magnitude of impulse is 2 kg m/s. The negative sign indicates the direction of impulse is opposite to the initial motion.

Mass of the man,

m = 50

kg.Height of the ladder,

h = 10

m.Time taken,

t = 20

seconds.Acceleration due to gravity,

g = 10

m/s

^{2}

.Work done by the man = Gain in potential energy =

m g h = 50 kg \times 10 {m/s}^{2} \times 10 m = 5000

J.Power generated by the man,

P = \frac{W}{t} = \frac{5000 J}{20 s} = 250

A fixed pulley is a wheel mounted on a rigid support, with a groove around its rim for a rope. When a force (effort) is applied downwards on one end of the rope, the load attached to the other end of the rope moves upwards.Diagram: A wheel with a rope passing over it, with the effort applied downwards on one side and the load attached upwards on the other side.How it makes work easier: It changes the direction of the effort, making it easier for the person to pull downwards than to lift upwards directly.Mechanical Advantage (MA) of a fixed pulley is 1.

M A = \frac{Load}{Effort} = 1

Length of the inclined plane,

L = 5

m.Mass of the object,

m = 100

kg.Height,

h = 2

m.Acceleration due to gravity,

g = 10

m/s

^{2}

.Load (weight of the object)

= m g = 100 kg \times 10 {m/s}^{2} = 1000

N.Mechanical Advantage (MA) of an inclined plane

= \frac{L}{h} = \frac{5 m}{2 m} = 2.5

.The person exerts a force (effort) of

F_{e f f o r t} = 400

N.Since

M A = \frac{Load}{Effort}

, we have

2.5 = \frac{1000 N}{400 N} = 2.5

.In an ideal frictionless inclined plane,

M A = \frac{Load}{Effort} = \frac{L}{h}

.Here, calculated MA from dimensions is 2.5, and MA from forces is

\frac{1000}{400} = 2.5

.This implies that the inclined plane is frictionless, or the frictional force is negligible. If there were friction, the actual mechanical advantage (Load/Effort) would be less than the ideal mechanical advantage (

L / h

). Since they are equal, we can assume it is frictionless.

A lever is a rigid bar that pivots around a fixed point called a fulcrum. It is used to multiply force or change its direction to perform work.Principle of a Lever: A lever works on the principle of moments. For a lever to be balanced, the clockwise moment must be equal to the anti-clockwise moment. Mathematically, Effort

\times

Effort Arm = Load

\times

Load Arm.Diagram: A rigid bar, a fulcrum point, effort applied at one end, and load at the other end or in between.

A lever is a rigid bar that pivots around a fixed point called a fulcrum.Examples and classification:Class I Lever: Fulcrum is between the effort and the load. Examples: Seesaw, scissors, crowbar, balance scale.Class II Lever: Load is between the fulcrum and the effort. Examples: Wheelbarrow, nutcracker, bottle opener.Class III Lever: Effort is between the fulcrum and the load. Examples: Tweezers, human forearm, broom, fishing rod.

Mass of the man,

m = 50

kg.Number of steps

= 50

.Height of each step

= 20

= 0.2

m.Total height climbed,

h = 50 \times 0.2

= 10

m.Time taken,

t = 25

seconds.Acceleration due to gravity,

g = 10

m/s

^{2}

.Work done by the man = Gain in potential energy =

m g h = 50 kg \times 10 {m/s}^{2} \times 10 m = 5000

J.Power of the man,

P = \frac{W}{t} = \frac{5000 J}{25 s} = 200

Mass of the cricket ball,

m = 200

= 0.2

kg.Initial velocity,

u = 20

m/s.Final velocity,

v = 0

m/s (brought to rest).Time taken,

Δ t = 0.05

seconds.Impulse

= Δ p = m v - m u = 0.2 kg \times (0 m/s - 20 m/s) = 0.2 \times (- 20)

kg m/s

= - 4

kg m/s.The magnitude of impulse is 4 kg m/s.Average force,

F_{a v g} = \frac{Impulse}{Δ t} = \frac{- 4 kg m/s}{0.05 s} = - 80

N.The average force applied by the fielder is 80 N in the direction opposite to the ball's motion.

Total mass of the family and car,

M = 200

+ 1000

= 1200

kg.Initial speed,

v = 72

km/h.Convert speed to m/s:

v = 72 \times \frac{1000}{3600} = 72 \times \frac{5}{18} = 4 \times 5 = 20

m/s.Kinetic energy just before brakes were applied,

K = \frac{1}{2} M v^{2} = \frac{1}{2} \times 1200 kg \times (20 m/s)^{2}

K = 600 \times 400 J = 240000

J.Time taken to stop,

t = 5

seconds.Final velocity,

v_{f} = 0

m/s.Change in kinetic energy,

Δ K = K_{f} - K_{i} = 0 - 240000

= - 240000

J.Work done by the brakes

= Δ K = - 240000

J.Average force exerted by the brakes,

F_{a v g} = \frac{Work done}{Displacement}

. We need to find the displacement first.Using

v_{f} = v_{i} + a t

0 = 20 m/s + a \times 5 s ⟹ a = - 4

m/s

^{2}

.Using

v_{f}^{2} = v_{i}^{2} + 2 a s

0^{2} = (20 m/s)^{2} + 2 \times (- 4 {m/s}^{2}) \times s

0 = 400 - 8 s ⟹ 8 s = 400 ⟹ s = 50

m.Average force,

F_{a v g} = \frac{Work done}{Displacement} = \frac{- 240000 J}{50 m} = - 4800

N.The average force exerted by the brakes is 4800 N in the direction opposite to the motion.

Download as PDF opens your browser's print dialog: choose Save as PDF as the printer.

Work, Energy, and Simple Machines — Important Questions

Question bank

Answers

All Class 9 Science chapters