4.) A curling stone is raised to the top of an inclined ice ramp with a height (

Question

4.) A curling stone is raised to the top of an inclined ice ramp with a height (vertical dimension) of 5.12 m and a width (horizontal dimension) of 19.6 m as shown above.

a) If it slides down the ramp without friction, how far does it then slide along the horizontal ice in front of the ramp, where the coefficient of friction is 0.0307?

b) Suppose the coefficient of static friction between a curling stone and an ice ramp is 0.0381. What is the steepest angle the ramp could make with the horizontal, before the stone began to slide?

Explanation / Answer

a) Ei = Ef
PE = kE
mgh = (1/2)mv^2
(9.8)(5.12) = (1/2)v2
v = 10 m/s
Through Newton's Second Law:
F = ma
ma = mu mg
a = (0.0307)(9.8)
a = 0.30086 m/s2
Now, using a kinematics equation, you can find thedistance:
vf^2 =vi^2 + 2a(x-xo)
0 = (10)^2 + 2(-0.30086)(x - 0)
x = 166.19 m -----------------------------------------
b) Fnet= ma
ma = Fg - Fs
ma = mg sin theta - mus mg cos theta
a = g sin theta - mus g cos theta
0 = (9.8) sin theta - (0.0381)(9.8) cos theta
tan theta = 0.0381
theta = 2.18 degrees
ma = mg sin theta - mus mg cos theta
a = g sin theta - mus g cos theta
0 = (9.8) sin theta - (0.0381)(9.8) cos theta
tan theta = 0.0381
theta = 2.18 degrees

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