Work And Energy Test

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Work And Energy Test - 3

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Question 1
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A person carrying a load of 50 kg on his head moves 10 m on a straight horizontal road with acceleration of 2 m/s². The amount of work done by him is

A
500 joules

B
50 9.8 10 joules

C
zero

D
1000 joules

Solution

Resultant force acting on the person in the horizontal direction, F = ma

F = 50 x 2 = 100 N

W = Fs = 100 x 10 = 1000 joules
Question 2
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Work done by centripetal force during circular motion is

A
(mv²/r)

B
zero

C
mv²/r²

D

Solution

Here, the centripetal force is provided by the tension in the string. The angle between the tension in the string and the displacement of the mass during the circular motion is 90°. So, the work done by the centripetal force is zero.
Question 3
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The graph below represents the relation between displacement 'x' and force 'F'. The work done in displacing an object from x = 8 m to x = 14 m is approximately

A
15 J

B
40 J

C
8 J

D
16 J

Solution

Area under the F-x graph gives the work done.
We can take the shape in the given graph approximately as right triangle (shown as dotted line).
Area of the triangle gives the work done.
Question 4
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A truck and a car are moving on a smooth, level road such that the kinetic energy associated with them is the same. Brakes are applied simultaneously to both of them such that equal retarding force is produced in both. Which one will cover greater distance before it stops?

A
Car

B
Truck

C
Both will cover the same distance

D
Nothing can be decided

Solution

According to work-energy theorem, Work done = Change in the kinetic energy

W = F_s = F_f - K_i

As the change in kinetic energy is the same, the retarding force (F) will be the same.
So, the distance travelled by the car is the same as that travelled by the truck.
Question 5
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A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is W₁ and in the second case is W_2.
Which of the following relations between W₁and W₂ is correct?

A
W₂ = W₁

B
W₂ = 2W₁

C
W₂ = 3W₁

D
W₂ = 4W₁

Solution

In case of the stretched spring,
Work done = Change in the elastic potential energy

Work done in the first case,
Work done in the second case,

Hence, W₂ = 4W₁
Question 6
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A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

A
t^1/2

B
t^3/4

C
t^3/2

D
t²

Solution

Assume that the body is initially at rest, and at any time let the velocity of the given body is v, as the power is constant, then

where
As the distance moved by the given body,
Question 7
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The work required to increase the velocity of a particle from 18 km/hr to 72 km/hr, if the mass of the particle is 2 kg, is

A
275 J

B
225 J

C
15 J

D
375 J

Solution

Initial velocity of the particle is

Final velocity of the particle is

Using work energy theorem
Question 8
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A stone of mass 500 gm is dropped from a certain height. When it is exactly at the midpoint of its free fall, the kinetic energy possessed by it is 800 J. What is the height from which it is dropped? (Take acceleration due to gravity of earth as 10 ms^-2)

A
320 m

B
160 m

C
80 m

D
240 m

Solution

Let the height from which the stone is dropped is 'h'.

At height , the kinetic energy possessed by the stone K = 800 J
Potential energy possessed by the stone U = mg
Using conservation of energy,
Total energy of the stone at height h = Total energy at height
Question 9
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If the momentum of a body is increased to 3 times its initial momentum, then by how much its kinetic energy will increase above its initial value which was 100 J?

A
200 J

B
300 J

C
900 J

D
800 J

Solution

As the kinetic energy K =
Final momentum p' = 3p

Hence the final kinetic energy,

K' = 9K' = 9K = 900 J

Change in the kinetic energy

K = K' - K = 900 - 100 = 800 J
Question 10
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Let us assume that you have lifted a suitcase from the floor and kept it on a table. The work done by you on the suitcase depends on

A
the path taken by the suitcase

B
the time taken by you in doing so

C
the weight of the suitcase

D
your weight

Solution

Wok done against gravity is independent of the path followed, and it is equal to the change in the potential energy.
Change in the potential energy = mgh, hence work done depends upon mg (weight of the suitcase).
Question 11
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If the kinetic energy of a body increases by 300%, then by what percentage shall the linear momentum of body increase?

A
200%

B
100%

C
150%

D
300%

Solution

As the kinetic energy K =
p =
Final kinetic energy K' = K +

p' =
Hence, we have

So, the final momentum p' = 2p

Percentage change in the momentum is
%
Question 12
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In a toy pistol, spring of spring constant 1000 N/m is compressed to 4 cm and released, which hits the ball of 10 grams initially at rest. Initial speed of the ball is

A
m

B
2m

C
3m

D
4m

Solution

Here, the elastic potential energy of spring is converted into the kinetic energy of the ball.

v =
Question 13
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A block is moving on a rough horizontal surface with constant velocity, then

A
the work done by the applied force and friction is negative

B
the work done by the frictional force is positive and normal reaction is negative

C
the work done by the applied force is positive and the frictional force is negative, and work done by the weight is zero

D
the work done by the gravitational force and the normal reaction is positive, and work done by the frictional force is negative

Solution

Here the angle between the direction of the motion and the applied force is zero; therefore, work done by the applied force is positive.
The directions of motion and the frictional force are opposite to each other; therefore, work done by the applied frictional force is negative.
The angle between the direction of motion and the weight is 90° and angle between the direction of motion and the normal reaction is 90°, so the work done by the both weight and normal reaction is zero.
Question 14
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An object initially at rest starts moving under the action of the force which varies with position as F = x, work done by this force in moving the object from x = 0 to x = 2 m is

A
2 J

B
1 J

C
4 J

D
6 J

Solution

As force F = x, on plotting the graph of the force and the position, we have

Work done = Area under the F-x graph (Area of the shaded portion)

W =
Question 15
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A spring 40 mm long is stretched by the application of a force. If 10 N force is required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is:

A
8 J

B
23 J

C
68 J

D
84 J

Solution

As F = kx so the spring constant k =
k = = 10⁴ N/m
Work done in stretching the spring by 40 mm
W = kx'²
W = × 10⁴ × (40 × 10^-3)² = 8 J
Question 16
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A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when length makes an angle of 60^o to the vertical, is:

A
2 m/s

B
m/s

C
m/s

D
2.5 m/s

Solution

Potential energy of the bob at point B
PE = mg(l - l cosθ)
From law of conservation of energy, we have
Total energy at point A = total energy at point B
m(3)² = mv² + mgl(1 - cosθ)

9 = v² + 2gl(1 - cos60°)
9 = v² + 2 × 10 × 0.5
v²= 4 m/s
v = 2 m/s
Therefore, the speed of bob is 2 m/s
Question 17
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Two springs of spring constant 1500 N/m and 3000 N/m are stretched with the same force. They will have the potential energies in the ratio of:

A
2 : 1

B
1 : 4

C
4 : 1

D
1 : 2

Solution

Potential energy stored in streched spring is

Spring force

Hence potential energy
Question 18
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A man carries 10 bricks to the top of a building by going up along a vertical spiral staircase in 5 minutes. A boy carries 10 bricks to the top of the building by going up a slanting ladder in 10 minutes, then

A
the man does more work than the boy

B
the boy does more work than the man

C
none of them does any work

D
both of them do the same work

Solution

As the gravitational field is a conservative field, work done by the gravitational force or the external force is independent of the path followed. It depends only on the initial and final positions.
Here, work done by the external force is equal to the change in the potential energy.
Change in potential energy in both cases is the same, so work done is the same.

However, time taken is different.
Hence, power will be different.
Question 19
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The engine of a bus of mass 5,000 kg accelerates the bus from 2 m/s to 20 m/s in 120 seconds. The power expended by the bus is

A
8250 W

B
8.25 W

C
82.5 W

D
825 W

Solution

As

Work done = change in the kinetic energy
, where v is the final

velocity, u is the initial velocity

W = 396 x 2500 = 990 kJ

Power =
Question 20
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An object of mass m is dropped from height 'H' and it reaches the ground in time 'T'. Study the following graphs and then pick the correct option. (v is the speed of the object at any instant and h is the height of the object at any instant)
(1).
(2).

(3).
(4).

A
Only 1

B
Only 1 and 3

C
Only 1, 3 and 4

D
Only 2, 3 and 4

Solution

For free fall, a = g
Using v = u + at
u = 0,
So, we can say that speed of the object is directly proportional to time. Graph 1 is incorrect and graph 2 is correct. Using conservation of energy, total mechanical energy of the object will remain conserved.

So, graphs 3 and 4 are also correct.