Question

A 190-g block is pressed against a spring of force constant 1.20 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0

Answer 1

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 = (1200 N/m)(0.100 m)^2 / [2(0.190 kg)(9.81 m/s^2)(sin60.0^o)]
L = 3.717 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 = (1200 N/m)(0.100 m)^2 / [2(0.190 kg)(9.81 m/s^2)((0.380)*cos60.0^o + sin60.0^o)]
L = 3.0483 m