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
If the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p2/2M. This must be equal to the maximum potential energy of the spring so that we have
p2/2M = ½ KL2, 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 = –6x3 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 (–6x3)dx = –6[x4/4] 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 v1 and v2 are the initial velocities of the man and boy respectively, we have
½ mv12 = ½ (m/2)v22
Therefore, v2 = v1√2.
On changing the speeds, we have
½ m(v1+2)2 = ½ (m/2)(v2+x)2
On substituting for v2 (=v1√2), the above equation simplifies to
(v1+2)2 = ½ (v1√2+x)2 from which x = 2√2 ms–1, given in option (b).
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