Work, Energy and Power Test

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Work, Energy and Power Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

How should the force applied on a body be varied with velocity to keep the power of force constant?

A
Force should be inversely proportional to the square root of the velocity of the body

B
Force should be inversely proportional to the velocity of the body

C
Force should be directly proportional to the velocity of the body

D
Force should not be varied. It should remain constant with the velocity

Solution

We know that power (P) = force (F) x velocity (v).
Since the body is moving with acceleration,v changes and as a result of that P also changes,F being constant.
For P to be constant, F v = constant or F ∝ 1/v i.e
Force should be inversely proportional to the velocity of the body.
Thus,as v increases, F should decrease to keep P constant.
Question 2
1 / -0

An engine generates a power of 10 kW. In how much time will it raise a mass of 200 kg through a height of 40m.

A
4 sec

B
10 sec

C
5 sec

D
8 sec

Solution

Power = work done / time
work = force x distance
so power = (force x distance) / time
time = (force x distance) / power
= (2000 N x 40m) / 10000W = 8 seconds
Question 3
1 / -0

A person 'A' pushes a body of 5 kg placed on a rough surface of co-efficient of friction 0.3 by a distance of 4 m in 15 seconds. Another person 'B' pushes the body of 7 kg on same surface by a distance of 3 m in the same time. Who has more energy and power?

A
A

B
B

C
Both A and B have developed equal power but B has more energy

D
Both A and B have developed equal power but A has more energy

Solution

For person A,
Mass of body, m₁= 5kg
Coefficient of friction, μ =0.3
So, The force of friction,
F1= μm₁g=0.3 X 5 X 10=15N
Displacement, S₁=4m
Therefore,
Work done, W₁=15X4=60J
Time taken, t₁=15sec
So, Power, P₁=60/15=4W
For Person B
Mass of body, m₂=7kg
Coefficient of friction, μ =0.3
So, The force of friction,
F₁= μm₂g=0.3 X 7 X 10=21N
Displacement, S2=3m
Therefore,
Work done, W₂=21X3=63J
Time taken, t₂=15sec
So, Power, P₃=63/15=4.2W
So person B has more energy and power.
Question 4
1 / -0

A body moves under a force such that momentum of the body at any instant . What is the rate of work done by the force?

A

B

C

D
zero

Solution

P = p Cos(t) i + p Sin(t) j
dP/dt = -p Sin(t) j + p Cos(t) i
Work done, W = P²/2m
Rate of work done = d/dt(P²/2m)
= 2P/²m x dP/dt
= P/m x dP/dt
= -p².Sin(t).Cos(t) + p².Sin(t).Cos(t)
=> Rate of work done = 0
Question 5
1 / -0

A man A of mass 80 kg runs up a staircase in 12 seconds. Another man of mass 60 kg runs up the same staircase in 11 seconds. The ratio of powers of A and B is :

A
11:12

B
9:11

C
12:11

D
11:9

Solution

According to question, 60 kg man runs up a staircase in 12 sec.
m₁ = 80 kg
t₁ = 12 sec
=> A 50 kg man runs up the same staircase in 11 sec:
m₂ = 60 kg
t₂ = 11 sec
=> Rate of work done is equals to power.
∴ P = E/ t and E = mgh
∴ P₁ = m₁gh / t₁
P₂ = m₂gh/ t₂
=> The ratio of Power or Rate of work done is:
P₁/P₂ = m₁gh / t₁ / m₂gh/ t₂
= m₁ * t₂ / m₂ * t₁
= 11 : 9
Thus, the ratio of rate of doing their work is 11 : 9.
Question 6
1 / -0

If momentum of an object is doubled, how does its kinetic energy change?

A
increases 4 times

B
increases 2 times

C
decreases 4 times

D
decreases 2 times

Solution

As the momentum, P = mv is doubled, we get the new velocity is 2v as mass can’t be increased. Thus if initial KE = ½ mv²
We get new KE = ½ m (2v)2
= 4 times the original one.
Question 7
1 / -0

An electric motor creates a tension of 4500 N in a hoisting cable and reels it in at the rate of 2 m/s. What is the power of the electric motor?

A
225W

B
15kW

C
9kW

D
900 W

Solution

F=4500N
v=2ms⁻¹
The power of motor is given by
P=F.v
P=45000×2
P=9000W=9kW
Question 8
1 / -0

From a waterfall, water is falling at the rate of 100 kg/s on the blades of a turbine. If the height of fall is 100 m, the power delivered to the turbine is

A
100kW

B
100W

C
1kW

D
10kW

Solution

Power = work done / time taken
or P = mgh / t
P = (m/t) gh = 100 x 10 x 100 = 100 kW
Question 9
1 / -0

Calculate the power developed when the force of 10 N is exerted on a body moving with velocity of 3.65 m/s.

A
36.5W

B
3.65W

C
365W

D
0.365W

Solution

Power =F.V
i.e, Power = force x velocity.
here, force (f)= 10N
and velocity (v) = 3.65 m/s
so, Power = f x v,
i.e, 10 x 3.65 = 36.5 w.
Question 10
1 / -0

A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time t is proportional to:

A
t^3/2

B
t^3/4

C
t^1/4

D
t^1/2

Solution

We know that P = F.v
= m.a.a.t
And with u= 0 we have s = ½ at²
Thus we get s = ½
Thus as P is constant and m is always the same
we get s is function of t^3/2

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Work, Energy and Power Test - 17

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Work, Energy and Power Test - 17

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Question 1

Force should be inversely proportional to the square root of the velocity of the body

Force should be inversely proportional to the velocity of the body

Force should be directly proportional to the velocity of the body

Force should not be varied. It should remain constant with the velocity

Solution

Question 2

4 sec

10 sec

5 sec

8 sec

Solution

Question 3

A

B

Both A and B have developed equal power but B has more energy

Both A and B have developed equal power but A has more energy

Solution

Question 4

zero

Solution

Question 5

11:12

9:11

12:11

11:9

Solution

Question 6

increases 4 times

increases 2 times

decreases 4 times

decreases 2 times

Solution

Question 7

225W

15kW

9kW

900 W

Solution

Question 8

100kW

100W

1kW

10kW

Solution

Question 9

36.5W

3.65W

365W

0.365W

Solution

Question 10

t3/2

t3/4

t1/4

t1/2

Solution

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t^3/2

t^3/4

t^1/4

t^1/2