A block of mass M = 10 kg is supported by a heavy rod on an incline of angle the

Question

A block of mass M = 10 kg is supported by a heavy rod on an incline of angle theta = 30 degree. The block is attached to a spring of length d = 2.2 meters which is relaxed (at equilibrium). The top of the spring is nailed to the incline. The support rod is removed and the block slides down along the incline until the spring stretches to length d+D, where its acceleration becomes zero. The spring constant is k=49.05 N/m. Assume that there is no friction anywhere in the system. How much does the spring stretch? Use g = 9.81 D =

Explanation / Answer

mgsintheta = k*x

x=stretch = 0.9989 m

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