Use figure 5 below. A block with mass m 2.00 kg is placed against a spring ona f
ID: 2040599 • Letter: U
Question
Use figure 5 below. A block with mass m 2.00 kg is placed against a spring ona frictionless incline with angle 0 30.0° (The block is not attached to the spring.) The spring, with spring constant k 1960 N/m, is compressed 20.0 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? (You will need to find the elastic potential energy which will be converted to KE of the block upon release, which will then become Gravitational potential energy at max height) (c) How far up the ramp will it move? (remember that the vertical distance upward is the side opposite of the angle and the distance up the ramp is the hypotenuse.)Explanation / Answer
(A) spring PE = k x^2 / 2
= (1960) (0.20^2) / 2
= 39.2 J
(B) change in PE = 39.2 J
(from energy conservation, elastic PE will convert into gravitational PE)
(C) changein PE = m g h
39.2 = 2 x 9.8 x d sin30
d = 4 m
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