Well, compute the spring energy at .34m,and at .14m. The difference is the amount of energy the spring has released. This energy goes to KE and change of GPE. You did not mention the slant angle, but GPE is gained (mgh)
spring energy released=KE+ mgh
of course spring energy= 1/2 k x^2, here the amount released is 1/2 k(.34^2-.14^2)
The drawing shows a block (m = 1.5 kg) and a spring (k = 300 N/m) on a frictionless incline. The spring is compressed by x0 = 0.34 m relative to its unstrained position at x = 0 m and then released. What is the speed of the block when the spring is still compressed by xf = 0.14 m?
1 answer
2 answers