A 2.00 kg block is placed against a spring on a frictionless 26° incline. The sp

Question

A 2.00 kg block is placed against a spring on a frictionless 26° incline. The spring, whose spring constant is 19.8N/cm, is compressed 17.6 cm and then released. What is the elastic potential energy of the compressed spring?

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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?

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How far along the incline is the highest point from the release point?

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Explanation / Answer

ELASTIC POTENTIAL ENERGY OF THE SPRING OF SPRING CONSTANT K and compressed to a distance 'x' can be find as -

U = 0.5 kx2

U = 0.5*19.8*17.6*17.6/100

U = 30.66 joule

when the block reached at highest point all the potential energy stored in the spring converts into potential energy of the block.

0.5 kx2 = mgh

mgh = 30.66

2*9.8*h = 30.66

h = 1.56 m

l = h/sin 26

l = 3.56 m

change in gravitational potential energy of the system-

initial PE of system = 0

final PE of system = mgh = 30.66

change = 30.66 - 0

= 30.66 joule

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