Please show all work :) A block with mass m = 1.07 kg is placed against a spring

Question

Please show all work :)

A block with mass m = 1.07 kg is placed against a spring on a frictionless incline with angle theta = 27.8 degree (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 29 N/cm, is compressed 21.8 cm and then released, (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point? Number Units Number Units Number Units

Explanation / Answer

a) E=kx^2/2 = 29*0.218^2/2 = 0.689 J

b) use energy conservation

Kx^2/2 = mgh --> h = 0.0657 m

c) sin(teta) = h/L --> L=h/sin(teta)=0.14 m

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