Question

A 20 kg box is attached to a compressed spring that has a spring constant of 300 N/m. the box is resting on a frictionless surface and the spring is compressed 30 cm.


A. what is the EPE of spring?


B. what will be the KE of the box when the spring expands back to its natural length?

C. how fast will the box be moving after the spring releases the box?

Answer 1

Answer:

320 n/m

Explanation:

becase

300+20=320

Answer 2

A. The EPE of the spring is given by u=$\frac{1}{2}*k*x^2$

      Therefore,

                  $\frac{1}{2}*300*0.3^{2}$ = 13.5 Joule

B. The KE of the box when the spring expands back to its natural length will be same as EPE due to conservation of energy, EPE is converted to KE.

C.The box will be moving after the spring releases the box is given by

           K.E=$\frac{1}{2} *m*v^{2}$

      as we know,

                     K.E=13.5 J , m=20kg

      Using this in the equation,

                  v=1.16 m/s