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.
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
To calculate the distance traveled by the shuffleboard disc after the push stops, we need to use the kinematic equation of motion:
d = (1/2)at^2 + v₀t
where:
d is the distance traveled
a is the acceleration
t is the time
v₀ is the initial velocity (zero in this case)
First, let's calculate the acceleration using Newton's second law:
Fnet = ma
The net force is the horizontal force applied minus the frictional force:
Fnet = F_applied - F_friction
Now, let's calculate the weight of the disc:
Weight = mass * gravitational acceleration (g)
Weight = 0.250 kg * 9.8 m/s² (convert grams to kg)
Next, let's calculate the frictional force:
F_friction = coefficient of friction * normal force
The normal force is equal to the weight of the disc due to the absence of vertical acceleration:
F_friction = (0.60) * Weight
Now, let's calculate the acceleration:
F_net = F_applied - F_friction
a = F_net / mass
After calculating the acceleration, we can find the distance traveled using the equation mentioned earlier.