A 2.00 kg is released on a 53.1degree incline, 4.0 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are static friction constant = 0.40, and kinetic friction constant= 0.20. The mass of the spring is negligible.
a)what is the speed of the package just before it reaches the spring.
b)what is the maximum compression of the spring?
c)the package rebounds back up the incline. How close does it get to its initial position?
The Attempt at a Solution
K1+U1+Wother = K2+U2
s is 4m, the distance it needs to get to the spring.
f= fk + fs
how do i get n? (n=wcos53.1??)
in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f=[-(0+mgh)-(0.5mv^2 + 0)]/4 = 7.06
to solve for v, we need h, is h simple sin53.1?
(*i made a diagram as well, but it isnt allowing me to post it...*)