Please provide answer AND an explantion on how you solved the problem. Thanks. A

Question

Please provide answer AND an explantion on how you solved the problem. Thanks.

A 225-g block is pressed against a spring of force constant 1.35 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0 degree to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions. (a) if the ramp exerts no friction force on the block m (b) if the coefficient of kinetic friction is 0.360 m

Explanation / Answer

a)

initial potential energy of spring = final gravitational potential energy of block

kx^2 / 2 = mgy

kx^2 / 2 = mgLsin

L = kx^2 / [2mgsin]

L = (1350 N/m)(0.100 m)^2 / [2(0.225 kg)(9.81 m/s^2)(sin60°)]

L = 3.53 m

b)

initial potential energy of spring = (work done by friction) + (final gravitational potential energy of block)

kx^2 / 2 = NL + mgy

kx^2 / 2 = mgLcos + mgLsin

L = kx^2 / [2mg(*cos + sin)]

L = (1350 N/m)(0.100 m)^2 / [2(0.225 kg)(9.81 m/s^2)((0.360)*cos60° + sin60°)]

L = 2.922 m

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