A block with mass m 2.00 kg is placed against a spring on a frictionless incline

Question

A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 28°. The block is not attached to the spring). The spring, with spring constant k = 19.8 N/cm, is compressed 20.2 cm and then released. What is the elastic potential energy of the compressed spring? Submit Answer Tries 0/15 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? Submit Answer Tries 0/15 How far along the incline is the highest point from the release point? Submit Answer Tries 0/15

Explanation / Answer

elastic PE = k x^2 /2

= (19.8 x 10^2 N/m) (0.202 cm)^2 / 2

= 40.4 J .........Ans

Work done by spring + Wortk done by gravity = change in KE

40.4 + W = 0 - 0

w = - 40.4

changein PE = - W = 40.4 J .......Ans

40.4 = m g d sin28

d = (40.4) / (2 x 9.8 x sin28)

d = 4.39 m

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