a) Ke at bottom= PE at the top= mgh
b) 1/2 m v^2=KE
c) spring PE= KE of ball hitting it
= 1/2 k x^2 solve for x
d) energy lost on ramp= .1*mg*SinTheta*2.0m
so subtract that from the initil energy mgh, and then set that equal to 1/2 m v^2 and solve for v.
A 1.20kg ball is rolled down a frictionless ramp whose top is 0.80m above a horizontal spring with ks = 800N/m. The ball reaches the horizontal just in front of the spring.
a.) What is the kinetic energy of the ball just before it contacts the spring?
b.) What is the velocity of the ball at that point using energy methods?
c.) After the ball contacts the spring, what distance will the spring be compressed from its equilibrium point?
d.) If the ramp is 2.00m long and has a coefficient of friction of μk = 0.10, use energy methods to find the speed of the ball at the bottom of the ramp.
1 answer
1 answer