The elastic potential enegry of a stretched spring is given by `E = 50 x^(2)`. Where `x` is the displacement in meter and and `E` is in joule, then the force constant of the spring is

Derive an expression for the potential energy of an elastic stretched spring.

A spring is compressed by 10 cm by a ball of mass 20 g . What will be the potential energy of a spring if the force constant of the spring is 20 N cm^(-1) .

A small trolley of mass 2.0kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation a speed of 600rpm, the length of stretched spring is 50cm. If the original length of the spring is 42cm, find the force constant of the spring.

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is :

The spring force is given by F=-kx , here k is a constant and x is the deformation of spring. The F-x graph is

If the potential energy of a spring is V on stretching it by 2 cm , then its potential energy when it is stretched by 10 cm will be

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:-

A mass 1 kg is attaced to the hook of a spring and the spring is suspended vertically from a ceiling (see Fig. 5.12a). The spring is displaced from its equilibrium position by a distance x. The spring constant of the spring is 2.0xx10^(2)" N"//"m" . Calculate the displacement x. Figure 5-12 (a) Mass of 1 kg attached to a spring is suspended vertically. (b) Spring displace from equilibrium position by a distance x.

If the potential energy of a spring is V on stretching it by 2cm , then its potential energy when it is stretched by 10cm will be