# Solved Examples

1.    A spring of spring constant 50 N/m is compressed from its natural position through 1 cm. Find the work done by the spring-force on the agency compressing the spring.
Sol.    The magnitude of the work is
1/2(kx2) =  1/2× (50 N/m) × (1 cm)2
= (25 N/m) × (1 × 10–2m )2  = 2.5 × 10–3 J
As the compressed spring will push the agency, the force will be opposite to the displacement of the point of application and the work will be negative. Thus, the work done by the spring-force is
–2.5 mJ.

2.    A particle of mass 20 g is thrown vertically upwards with a speed of 10 m/s. Find the work done by the force of gravity during the time the particle goes up.
Sol.    Suppose the particle reaches a maximum height h. As the velocity at the highest point is zero, we have
0 = u2 – 2gh
or,                    h = u2/2g
The work done by the foce of gravity is
–mgh  = –mg(u2/2g)  = –1/2(mu2)
= –(0.02 kg) × (10 m/s)2  = –1.0 J

3.    Two charged particles A and B repel each other by a force k/r2,where k is a constant and r is the separation between them. The particle A is clamped to a fixed point in the lab and the particle B which has a mass m, is released from rest with an intial separation r0 from A. Find the change in the potential energy of two-particle system as the separation increases to a large value. What will be the speed of the particle B in this situation?

Sol.    The situation is shown in figure (8.9). Take A  + B as the system. The only external force acting on the system is that needed to hold A fixed.(You can imagine the experiment being conducted in a gravity free region or the particle may be kept and allowed to move on  a smooth horizontal surface, so that the normal force balances the force of gravity). This force does no work on the system because it acts on the charge A which does not move. Thus , the external forces do no work and internal forces; are conservative, the total mechanical energy must, therefore, remain constant . there are two internal forces ; FAB acting on A and FBA acting on B. The force FAB does no work because it acts on A which does not move. The work done by FBA as the particle B is taken away is ,

4.    A block of mass m slides along a frictionless surface as shown in the figure (8.10). If it is released from rest at  A , wha