Work Energy And Power MCQs for NEET — Physics Questions with Answers

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A force (F)=3i^+cj^+2k^ acting on a particle causes a displacement: (s)=4i^+2j^+3k^ in its own direction. If the work done is 6 J then the value of ‘c’ is 

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Explanation

W=(3i^+cj^+2k^).(4i^+2j^+3k^)=6Joule

W=12+2c+6=6c=6  

A force F=(5i^+4j^)N acts on a body and produces a displacement S=(6i^5j^+3k^)m. The work done will be 

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Explanation

W=F.s=(5i^+4j^).(6i^5j^+3k^)

=3020=10J

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

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Explanation

Mass of the hanging length=60200×4=1.2kgWork done in pulling the chain=potential energy of the hanging chain=mgh=1.2×10×60×10-22=3.6 J

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that 

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Explanation

When a force of constant magnitude which is perpendicular to the velocity of particle acts on a particle, work done is zero and hence change in kinetic energy is zero.

A force F=(5i^+3j^+2k^)N is applied over a particle which displaces it from its origin to the point r=(2i^j^)m. The work done on the particle in joules is 

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Explanation

W=F.r=(5i^+3j^+2k^).(2i^j^)=103=7J   

Two bodies of masses 1 kg and 5 kg are dropped gently from the top of a tower. At a point 20 cm from the ground, both the bodies will have the same 

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Explanation

Velocity of fall is independent of the mass of the falling body.

A particle moves under the effect of a force F = Cx from x = 0 to x = x1. The work done in the process is 

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Explanation

W0x1F.dx=0x1Cxdx=Cx220x1=12Cx12  

A cord is used to lower vertically a block of mass M by a distance d with constant downward acceleration g4. Work done by the cord on the block is 

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Explanation

The cord applies a constant downward force of (3M/4)g on the block to cause a downward acceleration of g/4. According to the work-energy theorem, the work done by the cord on the block is equal to the change in kinetic energy, which is (1/2)Mv^2 - (1/2)Mu^2 = (1/2)M(v^2 - u^2) = -(3Mgd/4), where u = 0.

Two springs have their force constant as k1 and k2(k1>k2). When they are stretched by the same force 

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Explanation

W=F22k

If both springs are stretched by same force then W1k

As k1>k2 therefore W1<W2

i.e. more work is done in case of second spring. 

The potential energy of a certain spring when stretched through a distance ‘S’ is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance ‘S’ will be:

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Explanation

12kS2=10J(given in the problem)

W= 12k(2S)2(S)2=3×12kS2 = 3 × 10 = 30 J  

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