Here is a simple question which is meant for gauging your understanding of the work-energy principle. This MCQ appeared in AIEEE 2005 question paper:

**The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is**

**(a) ML**

^{2}/K (b) zero (c) KL^{2}/2M (d) √(MK).LIf the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p^{2}/2M. This must be equal to the maximum potential energy of the spring so that we have

p^{2}/2M = ½ KL^{2}, from which p = L√(MK)., given in option (d).

Now consider the following MCQ which appeared in Kerala Medical Entrance 2006 question paper:

**The work done by a force F = ****–6x ^{3} **

**î newton, in displacing a particle from x = 4m to x =**

**–**

**2m is**

**(a) 360 J (b) 240 J (c) ****– ****240 J (d) ****– 360 J ****(e) 408 J **** **

This is a case of variable force (in the X-direction), the point of application of which is moved in the X-direction. The work done is therefore given by

W = ∫F.dx = _{4 }∫^{–}^{2} (–6x^{3})dx = –6[x^{4}/4] with x between limits 4 and –2.

Therefore, W = –(6/4)(16 – 256) = 360 joule, given in option (a).

The following MCQ appeared in Kerala Engineering entrance 2006 question paper:

**A running man has the same kinetic energy as that of a boy of half the mass. The man speeds up by 2**** ms**^{–1} and the boy changes his speed by ‘x’** ms**^{–1} so that the kinetic energies of the boy and the man are again equal. Then ‘x’ in **ms**^{–1} is

**(a) ****–**** 2√2 (b) + 2√2 (c) √2 (d) 2 (e) 1/√2 **** ** ^{}

If the mass of the man is ‘m’, the mass of the boy is m/2. If v_{1} and v_{2} are the initial velocities of the man and boy respectively, we have

½ mv_{1}^{2} = ½ (m/2)v_{2}^{2}

Therefore, v_{2} = v_{1}√2.

On changing the speeds, we have

½ m(v_{1}+2)^{2} = ½ (m/2)(v_{2}+x)^{2}

On substituting for v_{2} (=v_{1}√2), the above equation simplifies to

(v_{1}+2)^{2} = ½ (v_{1}√2+x)^{2} from which **x = 2√2 ms**** ^{–1}**, given in option (b).

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