WORK ENERGY AND POWER MCQ TEST 1

Question 1.

--------------product of force and dispalcement is called workdone.

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Cross

Dot

Explanation

Answer : Dot product

Question 2.

Workdone is a ...............physical quantity.

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Vector

Scalar

Explanation

Answer: Scalar

Work done is a dot product of vectors force and the displacement. Since, the dot product is a scalar quantity.

Question 3.

What is dimensional formula of workdone?

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[M²L²T¯²]

[ML²T¯²]

[MLT¯²]

None

Explanation

Answer: [ML²T¯²]

The dimensional formula of work done is [ML²T¯²].

Question 4.

The magnitude of work done by a force:

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depends on frame of reference

does not depend on frame of reference

cannot be calculated in non-inertial frames.

both (a) and (b)

Explanation

Answer : depends on frame of reference

The work done by a force is dependent on the frame of reference.

Question 5.

Work done by a force on an object is zero when :

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displacement is zero

force is zero

force and displacement are mutually perpendicular

All of these

Explanation

Answer:All of these

W=Fdcosθ

When d=0,W=0

When F=0,W=0

When θ=90°,W=0

Question:6

When the force acting on body retards the motion of body, the work done will be:

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negative

positive

zero

None

Explanation

Answer: Negative

Force is acting opsite of direction of velocity or displacement

The angle between force and displacement is 180°.

Question:7

Work is always done on a body when:

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a force acts on it

it moves through a certain distance

it experiences an increase in energy through a mechanical influence.

None of these

Click Here For Solution

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

According to work energy theorem work done on body =change in kinetic energy of the body

Question:8

A force of 50 N acts on a body and moves it through a distance of 4 m on a horizontal surface. What would be work done in Joules if the direction of force is at an angle of 60∘ to the horizontal surface.

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200J

150J

100 J

None

Explanationn

Answer:100j

Given:

=>Force (F) = 50 N

=>Distance (d) = 4 m

=>Angle θ= 60∘

=>Therefore, W=50×4×1/2

(since cos 60∘ = ½)

=>W=100 J

Question:9

Which of following is correct ?

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Workdone= (force)*(displacement)

Workdone= (force)*(component of displacement in direction of force)

Workdone= (displacement)*(component of force in direction of displacement)

All of these

Expanation

Answer:All of these

W=Fdcosθ

When θ=90°,W=Fd [Workdone= (force)*(displacement)]

W=F(dcosθ)= (force)*(component of displacement in direction of force)

W=d(Fcosθ)= (displacement)*(component of force in direction of displacement)

Question:10

If the units of length and force are increased three times, then unit of energy will

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becomes 6 times.

becomes 8 times.

becomes 9 times.

None

Explanation

Answer:becomes 9 times.

W=Fdcosθ

When F1=3F,d=3d

W1=3F*3dcosθ

W1=9Fdcosθ

W1=9W

Question:11

A forceF =3i +cj +2k N acting on a particle causes a displacement S =−4i +2j −3k m. lf the work done is 6 Joule, the value of c is:

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0

1

12

None

Explanation

Answer:12

Question:12

A body of mass 6kg is under a force which causes displacement in it given by S= t²/4 metres where t is time. The work done by the force in 2 seconds is:

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12J

9J

3J

None of these

Explanation

Answer:3J

Given,S= t²/4

v=ds/dt=2t/4=t/2

v1 at t=0=>0

v2 at t=2=>1

Workdone= change in KE =1/2*6*1-0=3J

Question:13

A particle moves from position r1=3i +2j −6k to position r2=14i +13j +9k under the force 4i +j +3k . Find the work done.

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10J

100J

150J

None

Explanation

Answer:100J

Question:14

The work done against gravity in taking 10kg mass at 1m height in 1s will be (in J)

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196J

49J

98J

None of these

Explanation

Answer:98J

Work done against gravity =mgh=10×9.8×1=98J

Question:15

A force acting on a particle varies with the displacement x as F=ax−bx². Where a=1 N/m and b=1 N/m². The work done by this force for the first one meter (F is in newtons, x is in meters) is :

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(3/6)J

(2/6)J

(1/6)J

None

Explanation

Answer:(1/6)J

Question:16

A force F=(3xi +4j ) Newton (where x is in metres) acts on a particle which moves from a position (2 m, 3 m) to (3 m, 0 m). Then the work done is :

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-4.5J

-7.5J

16J

None of these

Explanation

Answer:-4.5J

Question:17

The position of a particle varies with time as x = (t-2)², where x is in meters and t is in seconds. Calculate the work done during t =0 to t = 4s , if mass of the particle is 100g.

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0.7J

0.5J

0

None of these

Explantion

Answer:0

Question:18

The minimum work done in slowly pulling up a block of wood weighing 2kN for a length of 10m on a smooth plane inclined at an angle of 15° with the horizontal by a force parallel to the incline is : (sin15°=0.26,cos15°=0.96)

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4.36KJ

5.36KJ

3.36KJ

5.17KJ

Click Here For Solution

Weight of the block (W=mg)=2 kN = 2000 N

Length of the incline (d) = 10 m

Angle of the incline (θ) = 15°

Height raised =10sin15°

Work done = Change in potential energy of block

⇒W=mgh

⇒W= 2000*10sin15°= 5.17 KJ

Question:19

Unit of energy is:

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kwh

joule

electron volt

All of the above

Click Here For Solution

Answer:All of the above

Question:20

A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is

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Negative

Zero

Positive

None of these

Explanation

Answer:Zero

Workdone =0 because there no force acting on ball by the table.