Question

Hi, ive been trying to get this problem for a couple of days now but i have no idea of what to do.
A very slippery block of ice slides down a smooth ramp tilted at angle pheta.

a)The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom.

b) Evaluate your answer to part a) for ice released at a height of 30cm on ramps tilted at an angle of 20 degrees and at 40 degrees.

Answer 1

I bet the angle was theta :)

If there is no friction the kinetic energy at the bottom, (1/2)m v^2 will be equal to the change in potential energy in the slide mgh
(1/2) v^2 = gh
v = sqrt (2gh) remember that

b) The angle does not matter if h is the same !!!!
h = 0.30 meters, g = 9.81 m/s^2

Answer 2

1/2at^2