Question

A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a spring that is anchored to a vertical wall, such that the spring is parallel to the surface of the incline.
a. If the crate was connected to the spring at equilibrium length, and then allowed to stretch the spring until the crate comes to rest, determine the spring constant. Assume that the incline is frictionless and that the change in length of the spring is 2.13 m.
b. If there is friction between the incline and the crate, would the spring stretch more, or less than if the incline is frictionless? You must use concepts pertaining to work and energy to receive full credit.

Answer 1

Answer:

Explanation:

Change in length of spring = 2.13 m

Component of weight acting on spring = mg sinθ

so

mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.

Here x = 2.13

mg sin17 = k x 2.13

31 x 9.8 sin17 = k x 2.13

k = 41.7 N/m

b ) In case surface had friction , spring would have stretched by less distance .

It is so because , the work done by gravity in stretching down is stored as potential energy in  spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc  so spring stretches less.