When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored in the wire is `1/2xxstressxx`strain`xxvolume`. Show that it is equal to `1/2Mgl` where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining `1/2Mgl` energy go ?

What is potential energy ? Derive an equation for gravitational potential energy of a body of mass .m. at a height .h.. (AS_(1))

Three bodies of each of mass 2kg are situated on x-axis at distance 1m, 2m , 4m from origin. The resolving gravitational potential due to the system at the origin will be

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x)=kx^(2) . Where k is a positive constant. If the amplitude of oscillation is a, then its time period T is……………

The length of a metal wire is l_1 when the tension in it is T_1 and is l_2 when the tension is T_2 . The natural length of the wire is

An infinite number of bodies, each of mass 2 kg, are situated on at distance 1m, 2m, 4m, 8m from the origin. What will be the resultant gravitational potential due to this system at the origin.

Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particle with usual reference level?

The potential energy of a particle in a force field is : U= A/r^2-B/r^1 , where A and B are positive constant and r is the centre of the field. For stable equilibrium, the distance of the particle is :