Work And Energy(9th Grade > Physics ) Questions and answers for exam Preparation

9th Grade > Physics

WORK AND ENERGY MCQs

Total Questions : 131 | Page 3 of 14 pages

A loader applied $150 N$ of force to lift a stone and $1250 J$ of work was done by it. How far was the rock lifted during the application of force?

$9.3 m$
$10.3 m$
$8.3 m$
$7.3 m$

Discuss Question

Answer: Option C. ->

8.3 m

:
C
Given:
Work done,

W = 1250 J

Force,

F = 150 N

Let the distance the rock was raised be

s

.
Work done is the product of force and displacement of the body along the force.

W = F s

s = \frac{1250}{150} = 8. ¯ 3 m

Thus, the rock was lifted through a distance of

8.3 m

Question 22.

Hulk tried to move the wall of a fortress, but unfortunately he was unable to do it. But he applied a very large amount of force to do so, hence work done is positive.

True
False
$8.3 m$
$7.3 m$

Discuss Question

Answer: Option B. -> False
:
B
Work done by a force is defined as the product of force and displacement of the body along the direction of force. Since Hulk was unable to move the wall, the displacement is zero and irrespective of the large amount of force he applied, work done will be zero.

Question 23.

Moving train has complete mechanical energy in the form of potential energy.

True
False
Just after touching the ground floor
Just before touching the ground floor

Discuss Question

Answer: Option B. -> False
:
B

Moving train has velocity and will have only kinetic energy. A moving train will not gain any height and hence has no potential energy. Therefore, the given statement is wrong.

Question 24.

A child is moving on a skateboard at the speed of 2 $m s^{- 1}$ . After a force acts on him, his speed becomes 3 $m s^{- 1}$ . Find the work done by the external force on the child, given the mass of the child is 20 kg.

100 J
50 J
20 J
320 J

Discuss Question

Answer: Option B. -> 50 J
:
B

Let initial velocity of the child be $v_{1}$ and final velocity be $v_{2}$
Given, $v_{1} = 2 m s^{- 1}$ , $v_{2} = 3 m s^{- 1}$ and mass of the child $= 20 k g$ .
Since the child is in motion, he will have a kinetic energy of $\frac{m v^{2}}{2}$ , which will be increased after an application of force by an external agent on him.
Now, Applying work energy theorem ,
Work done by the external force on the child $= \frac{m v_{2}^{2}}{2} - \frac{m v_{1}^{2}}{2}$

$= \frac{20 \times (9 - 4)}{2}$

$= 10 \times 5 = 50 J$

Question 25.

An object is falling from a terrace of height $20 m$ . At what point does it have maximum kinetic energy?

$20 m$ height
$10 m$ height
Just after touching the ground floor
Just before touching the ground floor

Discuss Question

Answer: Option D. -> Just before touching the ground floor
:
D

When the object is at a height of $20 m$ above the ground, it has only potential energy.
Since the velocity of the object increases when it comes down, its kinetic energy will increase. When the object is just about to touch the ground, its potential energy will be zero and kinetic energy will be maximum.

Question 26.

Which of the following are forms of mechanical energy?

Potential energy
Kinetic energy
Nuclear energy
Solar energy

Discuss Question

Answer: Option A. -> Potential energy
:
A and B

Mechanical energy is the sum of energy that is possesed by an object due to its motion and due to its position.
The two forms of mechanical energy are:

1. Potential energy: The energy possessed by a body due to its position is the potential energy.

2. Kinetic energy: The energy possessed by a body by virtue of its motion is the kinetic energy.

Question 27.

A man lifts a luggage of $10 k g$ from the ground and puts it on his head $1.7 m$ above the ground. Calculate work done by him on the luggage. Take acceleration due to gravity $= 10 m s^{- 2}$ .

$150 J$
$170 J$
$180 J$
$200 J$

Discuss Question

Answer: Option B. ->

170 J

:
B
Given:
Mass,

m = 10 k g

,
Acceleration due to gravity,

g = 10 m s^{- 2}

,
Displacement,

s = 1.7 m

.
Force

F

is calculated as

F = m \times g = 10 \times 10 = 100 N

Work done by the man on luggage

= F o r c e \times d i s p l a c e m e n t

W = 100 \times 1.7 = 170 J

Hence, the required work done is

170 J

.
It can be noted that the work is done by the man on the luggage because he lifts it against the gravitational force of the earth.

Question 28.

An electric bulb consumes 7.2 kJ of electrical energy in 2 minutes. What is the power of the electric bulb?

$40 W$
$60 W$
$50 W$
$70 W$

Discuss Question

Answer: Option B. ->

60 W

:
B
Given, energy E

= 7.2 k J = 7.2 \times 10^{3} J

and time t

= 2 m i n u t e s = 2 \times 60 = 120 s

.
We know, power,

P = \frac{E n e r g y o r W o r k}{t} = \frac{7.2 \times 10^{3}}{120} = 60 W

∴

The power of electric bulb is

60 W

Question 29.

When a lift is going up, the potential energy of the person with respect to earth inside a lift remains the same.

True
False
$50 W$
$70 W$

Discuss Question

Answer: Option B. -> False
:
B

Potential energy depends on the position of an object with respect to a reference point. In this case, the earth is the reference point. A man on a lift going up, gains height with respect to the earth.
Thus, the potential energy of the person inside the lift increases.

Question 30.

Potential energy depends on:

Change in position of the object.
Velocity with which the object moves.
Mass of the object.
Time required to move the object.

Discuss Question

Answer: Option A. -> Change in position of the object.
:
A and C

Potential energy is the energy possessed by an object by virtue of its position.

Potential energy possessed by an object is given by,

PE = $m \times g \times h$ , where m - mass of the object, g - acceleration due to gravity and h - height to which the object is raised from ground level.