Two Kerala Medical Entrance 2007 Questions on Work and Energy

The following two questions appeared in Kerala Medical Entrance 2007 question paper which contained relatively simple questions:

(1) When a bullet is fired at a target, its velocity decreases by half after penetrating 30 cm in to it. The additional thickness it will penetrate before coming to rest is

(a) 30 cm (b) 40 cm (c) 10 cm (d) 50 cm (e) 20 cm

Questions similar to this one often find place in admission test question papers. This can be worked out using the work-energy principle or by using equations of one-dimensional motion. Let us use the work-energy principle:

Since the decrease in kinetic energy is equal to the work done against the retarding force in the target, we can write

½ m[v² – (v/2)²] = F×0.3, where ‘m’ is the mass of the bullet, ‘v’ is its initial velocity and F is the retarding force, which has to be assumed to be constant.

If ‘s’ is the additional distance penetrated (for the velocity to decrease to zero from the value v/2),

½ m(v/2)² = Fs

Dividing the first equation by the second, 3 = 0.3/s, from which s = 0.1 m = 10 cm.

[ If you use the equation of uniformly retarded linear motion, v² = u² –2as where ‘u’ and ‘v’ are the initial and final velocities respectively and ‘a’ is the retardation, we have the two equations,

(u/2)² = u² – 2a×0.3 and

0 = (u/2)² – 2as

The first equation is 3u²/4 = 2a×0.3.

The second equation is u²/4 = 2as.

Dividing the first one by the second, 3 = 0.3/s from which s = 01 m = 10 cm].

(2) A body constrained to move in the Y-direction is subjected to a force F = 2i + 15j + 6k newton. The work done by this force in moving the body through a distance of 10 m along the Y-axis is

(a) 100 J (b) ) 150 J (c) ) 120 J (d) ) 200 J (e) 50 J

The work (W) done by a force F in producing a displacement S is given by the scalar product of the vectors F and S.

Therefore, W = F.S = (2i + 15j + 6k) . 10j = 150 J

[The displacement vector is 10j since it is along the Y-axis].

You may work this out also by arguing that the component of the force along the direction of displacement of 10 m is 15 newton so that the work done is 15×10 = 150 J.