A spring (k=80n/m ) has an equilibrium length of 1.00m . The spring is compressed to a length of 0.50m and a mass of 2.1kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees 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 will the mass move before coming to rest? b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? c) Now the incline has a coefficient of kinetic friction. 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?
I think the equation is: .5mvi^2 + mgyi + .5 kxi^2 = .5 mvf^2 + mgyf + .5 kxf^2
The Attempt at a Solution
a) The initial and final velocities are zero. You don't need the final spring configuration because the block leaves the spring.
mgyi + .5kxi^2=mgyf
2.1(9.8)(.5sin41) + .5(80)(.5^2) = 2.1(9.8)y?
do I want y as the answer or x? and if I want x would y = xsin41?
b) The initial and final velocities are zero.
c) Use Work done by friction = Change in PE (again no change in KE)