Well, you have to give me all the first parts before I can do the last. However:
Vi = initial speed when push stops
m g = weight down
-.6 mg = friction force, only horizontal force left
m a = friction force = -.6 mg
so
a = -.6 g deacceleration
so
when does it stop?
v = Vi - .6 g t
t = Vi/(.6 g) time to stop
where does it stop?
d = Vi t -(1/2)(.6g)t^2
A shuffleboard disc has a mass of 250g. The coefficient of friction between the disc and the floor is .60. A player pushes the disc applying a horizontal force of 5.0 N for .35 s. How much does the disc weight? What is the frictional force between the disc and the floor. What rate does the disc accelerate while being pushed? Assuming the disc is at rest before being pushed, what is the discs velocity at the end of the push? How far does the disc travel after the push stops? I got everything but the last question. I have no idea how to get it. Please Can anyone help me.
Thank You.
1 answer
1 answer