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# WORK ENERGY AND POWER MCQ TEST 2

Work Engergy and Power 11 CBSE - MCQ test 1

Question 1.

A block of mass m is placed on elevator moving down with accelertion g/3.What would be value of workdone by normal reaction acting on the block if elevator moves distance h downwards .

mgh/3

-2mgh/3

2mgh/3

None of these

Explanation

Question 2.

Force F acts on the free end of string as shwon pulley system below,if weight W moves up slowly a distance h.What would be value of work done by force F?

Fh

2Fh

Fh/2

2Fh/g

Explanation

When W moves a distance h above and string at free end will move 2h below along the direction of force.

Work done=F*2h=2Fh

Question 3.

Below figure is showing position time graph of mass 2Kg. What would be workdone by force for movement of mass from t=0 to t=4 sec.

8j

0

4j

Can't detemine

Explanation

Dispalcement after t=4sec is 0.

Question 4.

A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly oppose to the motion.How much work does the cycle do on the road?

0J

2000J

-2000J

None

Explanation

Question 5.

Three forces of magnitude 14N,7N and 7N acting on a body in direction of vector ( 6i+2j+3k),(3i-2j+6k) and (2i-3j-6k)respevtively. The foreces remains constant while object moves from point A (2,-1,-3) to A (5,-1,1).What would be workdone on object.

150J

125J

0

75J

Explanation

Question:6

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used ?

1.3hp

1.4hp

1.5 hp

None

Explanation

P = F v = mgv = (500 kg) (9.8 m/s2) (0.2 m/s) = 980 W Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is P = 980 W = 980/746 hp = 1.3 hp.

Question:7

A body dropped from a height H reaches the ground with a speed of 1.2 √gH. Calculate the work done by air-friction.

a force acts on it

it moves through a certain distance

it experiences an increase in energy through a mechanical influence.

None of these

Explanation

Answer: It experiences an increase in energy through a mechanical influence.

Question:8

A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block.

μmg(dcos θ) /(cosθ - μsinθ)

μmgd /(cosθ + μsinθ)

μmg(dcos θ) /(cosθ + μsinθ)

None

Explanationn

Question:9

Static friction is variable.

True

False

Expanation

Value of static friction changes when the value of applied force changes.

Question:10

A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platform where they took the jump, the total work done by the man at the instant he lands is ....

0

300 J

600 J

None

Explanation

Given,
Mass of man = 60 Kg
Distance covered by him as he jumps up = 1 m

The formula of work done is
Work = force x displacement
W = F x d
Now Force "F" = mg
W = mg x d
W = 60 x 10 x 1
W = 600 J
Thus the work done is 600 J

Question:11

A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

7.2J

10J

3.6J

None

Explanation

Mass of the chain lying freely from the table=(Mass of chain*hanging length of chain)/Total length of chain

=(4*0.60)/2=1.2kg

The distance of center of mass of chain from the table= 0.5×0.6m=0.3m

Thus the work done in pulling the chain=mgh=1.2×10×0.3J=3.6J

Question:12

A body of mass 20 kg is initially at a height of 3 m above the ground. It is lifted to a height of 2 m from that position. Its increase in potential energy is:

120 J

90 J

392 J

None of these

Explanation

Given,Mass, m = 20kg

Height m, H = 3m

Potential Energy, (PE1) = mgH ,PE1 = 20 x 9.8 x 3 PE1 = 588 joules

Now Height (h) is (3m + 2m)=5m

(PE2) = mgh

(PE2) = 20 x 9.8 x 5 (PE2) = 980 Joules

Increase in Potential Energy = PE2 - PE1 = 980 - 588 = 392 Joules

Question:13

The kinetic energy acquired by a mass m in travelling distance d, the starting from the rest under the action of a constant force, is directly proportional to:

m

1/m

Does not depend on mass

Explanation

Work done by external forces = Change in kinetic energy Since particle starts from rest, its initial Kinetic Energy is 0. Therefore, K.E.=Fd It is given that force is constant. We see that Kinetic energy is independent of mass.

Question:14

A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is:

116 J

316 J

216 J

None of these

Explanation

Total energy of a body(E) = Kinetic Energy (KE) + Potential Energy (PE)
Kinetic Energy (KE):
The kinetic energy of an object can be calculated using the formula:
KE = (1/2) * m * v^2
KE = (1/2) * 10 kg * (2 m/s)^2
KE = 20 J (Joules)
Potential Energy (PE):
The potential energy of an object near the surface of the Earth can be calculated using the formula:
PE = m * g * h
PE = 10 kg * 9.81 m/s^2 * 2 m
PE = 196.2 J (Joules)
Now, let's calculate the total energy:
E = KE + PE
E = 20 J + 196.2 J
E = 216.2 J
So, the total energy of the body is 216 Joules(APPROXIMATE).

Question:15

Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

4:1

3:2

1:2

None

Explanation

We have, K.E=P²/2M
P1²/2M = P2²/2M
P1²/2*1 = P2²/2*4
P1²/P2²= 2/8
(P1/P2)² = 1/4
(P1/P2) = (1/2)

Question:16

A body of mass 10 kg is initially at a height of 20 m above the ground. It falls to a height of 5 m above the ground. Its potential energy in the new position is:

490J

500J

1000J

None of these

Explanation

PE = m * g * h
PE = 10 kg * 9.8 m/s² * 5 m
PE = 490 J (joules)
So, the potential energy of the body in its new position, 5 m above the ground, is 490 joules.

Question:17

A body of mass 100 kg falls from a height of 10 m. Its increase in kinetic energy is

1000 J

1800 J

9800 J

None of these

Explantion

When an object is at some height the total energy in it is in the form of potential energy. When the body falls from
a height below the potential energy gets converted into kinetic energy.
Mass of body(m) = 100 kg
Height(h)= 10m
g=9.8m/s²
Since kinetic energy is equal to loss of potential energy when it falls down,
Increase in kinetic energy= loss in PE=100*9.8*10= 9800J

Question:18

A gardener pushes a lawn roller through a distance of 20 m. If he applies a force of 20 kg wt with in a direction in blind at an angle 60 degree to the ground find the work done by the gardener.

4.36KJ

5.36KJ

3.36KJ

5.17KJ

Explanation

Force aplied, F = 20 kg-wt = 20×9.8N=196N
Since,force is applied at an angle 60° with the ground.
So, the horizontal component of the force that is responsible for displacement of the lawn roller= Fcos60°=196×0.5=98N
So, work done=98×20=1960J

Question:19

If force 𝐹⃗ = (𝑖̂+ 5𝑗̂+ 7𝑘̂) acts on a particle and displaces it through 𝑑⃗ = (6𝑖̂+ 9𝑘̂). Calculate the work done if the force is in newton and displacement is in metre

89 J

79 J

59 J

69 J

Explanation

Force F = i + 5 j + 7 k
Displacement s = 6 i + 9 k
Work done = F . s = ( i + 5 j + 7 k ) . ( 6 i + 0 j + 9 k ) = 6 +0 +63 = 69 J.

Question:20

A force F acting on an object varies with distance x as shown in figure below. The force is in newton and distance x is in metre. The work done by the force in moving the object from x = 0 to x = 6 m is:

12.5 J

13.5 J

0

None of these

Explanation