Work Energy and Power Test

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Work Energy and Power Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is \(mv^2\)/r and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?

A
\(\frac{mv^2}{\pi r^2}\)

B
Zero

C
\(\frac{mv^2}{r^2}\)

D
\(\frac{\pi r^2}{mv^2}\)

Solution

Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.

W = \(\overrightarrow{F}.\overrightarrow{s}\) = F s cos \(\theta\) = F s cos(90°) = 0
Question 2
1 / -0

If the unit of force and length each be increased by four times, then the unit of energy is increased by

A
16 times

B
8 times

C
2 times

D
4 times

Solution

Work = (Force) × (Displacement) [length]

If unit of force and length be increased by four times then the unit of energy will increase by 16 times.
Question 3
1 / -0

A woman pushes a wall and fails to displace it. She does

A
Negative work

B
Positive but not maximum work

C
No work at all

D
Maximum work

Solution

No displacement is there.
Question 4
1 / -0

A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joules, the angle which the force makes with the direction of motion of the body is

A
0°

B
30°

C
60°

D
90°

Solution

W = Fs cos \(\theta\)

⇒ cos\(\theta\) = \(\frac{W}{Fs} = \frac{25}{50} = \frac{1}{2}\)

\(\theta\) = 60°
Question 5
1 / -0

A body of mass m kg is lifted by a woman to a height of one metre in 30 sec. Another woman lifts the same mass to the same height in 60 sec. The work done by them are in the ratio

A
1:2

B
1:1

C
2:1

D
4:1

Solution

Work done does not depend on time.
Question 6
1 / -0

A 50kg man with 20kg load on his head climbs up 20 steps of 0.25m height each. The work done in climbing is

A
5J

B
350J

C
100J

D
3430J

Solution

Total mass = (50 + 20) = 70 kg

Total height = 20 × 0.25 = 5m \ Work done = mgh

= 70 × 9.8 × 5

= 3430 J
Question 7
1 / -0

A girl starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by her

A
Zero

B
Positive

C
Negative

D
Nothing can be said

Solution

As surface is smooth so work done against friction is zero. Also the displacement and force of gravity are perpendicular so work done against gravity is zero.
Question 8
1 / -0

A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about

A
0.1 joule

B
0.2 joule

C
0.3 joule

D
0.5 joule

Solution

DP.E. = \(\frac{1}{2}k(x_2^2 - x_1^2) \)

= \(\frac{1}{2} \times 10[(0.25)^2 - (0.20)^2]\)

= 5 x 0.45 x 0.05

= 0.1 J
Question 9
1 / -0

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

A
84 J

B
68 J

C
23 J

D
8 J

Solution

Here,

k = \(\frac{F}{x} = \frac{10}{1 \times 10^{-3}}\) = \(10^4\) N/m

w = \(\frac{1}{2} kx^2\)

= \(\frac{1}{2} \times 10^4 \times (40 \times 10^{-3})^2\)

= 8 J
Question 10
1 / -0

Tripling the speed of the motor car multiplies the distance needed for stopping it by

A
3

B
6

C
9

D
Some other number

Solution

S ∝ \(u^2\) i.e. if speed becomes three times then distance needed for stopping will be nine times.
Question 11
1 / -0

An electric motor exerts a force of 40 N on a cable and pulls it by a distance of 30 m in one minute. The power supplied by the motor (in Watts) is

A
20

B
200

C
2

D
10

Solution

P = Fv

P = \(\frac{Fs}{t} = 40 \times \frac{30}{60}\)

= 20 W
Question 12
1 / -0

A weight lifter lifts 300 kg from the ground to a height of 2 meter in 3 second. The average power generated by him is

A
5880 watt

B
4410 watt

C
2205 watt

D
1960 watt

Solution

P = \(\frac{Workdone}{Time}\)

= \(\frac{mgh}{t}\)

= \(\frac{300 \times 9.8 \times 2}{3}\)

= 1960 W
Question 13
1 / -0

Assertion: Work done by friction on a body sliding down an inclined plane is positive.

Reason: Work done is greater than zero, if angle between force and displacement is acute or both are in same direction.

(A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

(C) If assertion is true but reason is false.

(D) If the assertion and reason both are false.

(E) If assertion is false but reason is true.

A
A

B
B

C
C

D
D

Solution

When a body slides down on inclined plane, work done by friction is negative because it opposes the motion (θ = 180° between force and displacement) If θ < 90° then W = positive because W = F. s. cosθ
Question 14
1 / -0

A ball hits the floor and rebounds after inelastic collision. In this case

A
The momentum of the ball just after the collision is the same as that just before the collision

B
The mechanical energy of the ball remains the same in the collision

C
The total momentum of the ball and the earth is conserved

D
The total energy of the ball and the earth is conserved

Solution

By the conservation of momentum in the absence of external force total momentum of the system (ball + earth) remains constant.
Question 15
1 / -0

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to \(\frac{-K}{r^2}\), where K is a constant. The total energy of the particle is

A
\(\frac{K}{2r}\)

B
\(-\frac{K}{2r}\)

C
\(-\frac{K}{r}\)

D
\(\frac{K}{r}\)

Solution

\(\frac{mv^2}{r} = \frac{K}{r^2}\) K.E. = \(\frac{1}{2}mv^2 = \frac{K}{2r}\)

U = \(- \int_{\infty}^r F.dr\)

= \(- \int_{\infty}^r \Big(- \frac{K}{r^2} \Big)dr = - \frac{K}{r}\)

E = K.E. + P.E. = \(\frac{K}{2r} - \frac{K}{r} = - \frac{K}{2r}\)
Question 16
1 / -0

A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then

A
Lorry will come to rest in a shorter distance

B
Car will come to rest in a shorter distance

C
Both come to rest in a same distance

D
None of the above

Solution

Stopping distance = \(\frac{kinetic\;energy}{retarding\;force}\)

s = \(\frac{1}{2}\frac{mu^2}{F}\)

If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance.
Question 17
1 / -0

The work done against gravity in taking 10 kg mass at 1m height in 1sec will be

A
49 J

B
98 J

C
196 J

D
None of these

Solution

Work done = mgh = 10 × 9.8 × 1

= 98J
Question 18
1 / -0

In an explosion, a body breaks up into two pieces of unequal masses. In this

A
Both parts will have numerically equal momentum

B
Lighter part will have more momentum

C
Heavier part will have more momentum

D
Both parts will have equal kinetic energy

Solution

Both parts will have numerically equal momentum and lighter part will have more velocity.
Question 19
1 / -0

If force and displacement of particle in direction of force are doubled. Work would be

A
Double

B
4 times

C
Half

D
\(\frac{1}{4}\) time

Solution

Work = Force × Displacement

If force and displacement both are doubled then work would be four times.

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 12

Result

Work Energy and Power Test - 12

Score

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SHARING IS CARING

Question 1

\(\frac{mv^2}{\pi r^2}\)

Zero

\(\frac{mv^2}{r^2}\)

\(\frac{\pi r^2}{mv^2}\)

Solution

Question 2

16 times

8 times

2 times

4 times

Solution

Question 3

Negative work

Positive but not maximum work

No work at all

Maximum work

Solution

Question 4

0°

30°

60°

90°

Solution

Question 5

1:2

1:1

2:1

4:1

Solution

Question 6

5J

350J

100J

3430J

Solution

Question 7

Zero

Positive

Negative

Nothing can be said

Solution

Question 8

0.1 joule

0.2 joule

0.3 joule

0.5 joule

Solution

Question 9

84 J

68 J

23 J

8 J

Solution

Question 10

3

6

9

Some other number

Solution

Question 11

20

200

2

10

Solution

Question 12

5880 watt

4410 watt

2205 watt

1960 watt

Solution