A spring (80 N/m ) has an equilibrium length of 1.00 m. The spring is compressed

Question

A spring (80 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 with respect to the horizontal. The spring is then released.

a)If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

b)If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

c)Now the incline has a coefficient of kinetic friction k. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction k?

Explanation / Answer

when the spring is released, the spring potential energy will be converted to kinetic energy plus an increase in potential energy. At the stopping point up the ramp, the kinetic energy has also been converted to potential energy so we can essentially ignore the fact that kinetic energy existed in the system.
PE = PS
mgh = ½kx²

with our origin set at the maximum spring compression point
h = kx²/2mg

then convert to distance up the slope by geometry
d = h / sin41

d = 0.71 m

2) is the same sort of question with a twist. The spring is attached so the energy of spring compression is converted to energy of spring extension plus an increase in potential energy

PS+ PE = PS
PE = PS- PS
mgh = ½kx²- ½kx²
mgh = ½k(x²- x²)

from geometry we know that
h = dsin
and
x= d - x

mgdsin = ½k(x²- (d -x)²)
mgdsin = ½k(x²- d² + 2dx² - x²)
2mgdsin / k = - d² + 2dx²
d² + 2mgdsin / k - 2dx²=0
d² + d[2mgsin / k - 2x²] =0
d(d + [2mgsin / k - 2x²]) =0

so either
d = 0 which is irrelevant because the spring has not been released
or
d = -2mgsin / k - 2x²which will be our answer

d = 0.65 m

3) Now the energy of spring potential will be converted into increased potential energy and work within the system

PE + U = PS
mgh + Fd = ½kx²

here
d = x
and
h = dsin
and
F = kinetic friction force

the normal force of the ramp on the block is
Fn = mgcos
so
F = kmgcos

mgdsin + kmg(cos)d = ½kd²
mgsin + kmgcos = ½kd
sin + kcos = kd/2mg
k = [kd/2mg - sin] /cos

k = 0.36

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