Question

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg. The spring is released and the object moves along a frictionless surface when it reaches a small embankment. If the speed of the object is 2.45 m/s at location A, what is the embankment height h

Answer 1

Given :

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.

To Find :

The embankment in the height.

Solution :

Since no external force is acting in the system, therefore total energy will be conserved.

Initial kinetic energy of the object = Energy stored in spring

`K.E _i = (kx^2)/(2)\\\\K.E_i = (1730* 0.136^2)/(2)\\\\K.E_i = 16\ J`

Also, initial potential energy is 0.

Now,

`K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = (1.72* 2.45^2)/(2)+ mgh\\\\mgh =16 - 5.16\\\\h = (16 - 5.16)/(1.72 * 9.8)\\\\h = 0.64\ m`

Therefore, the embankment height is 0.64 m.