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.
I replied below.
To find the distance the shuffleboard disc travels after the push stops, you can use the equation of motion:
d = v0t + (1/2)at^2
Where:
d = distance traveled (unknown)
v0 = initial velocity (unknown)
t = time (given as 0.35 s)
a = acceleration (unknown)
We can determine the initial velocity using the equation:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (unknown)
a = acceleration (from the previous question)
Since the disc was at rest before being pushed, the initial velocity (u) is 0. Therefore, the equation simplifies to:
v = at
Now, let's calculate the acceleration (a) using the force applied and the formula:
F = ma
Where:
F = applied force (5.0 N)
m = mass of the disc (250g or 0.25 kg)
We also need to determine the weight of the disc. Weight is given by the formula:
Weight = mass * acceleration due to gravity
Where:
mass = 0.25 kg
Now, let's calculate the weight and acceleration:
Weight = 0.25 kg * 9.8 m/s^2
Weight = 2.45 N
Next, we can calculate the acceleration:
F = ma
5.0 N = 0.25 kg * a
a = 20 m/s^2
Plugging the values into the equation for v (final velocity):
v = at
v = 20 m/s^2 * 0.35 s
v = 7 m/s
Now we can use the equation for d (distance traveled) with the known values:
d = v0t + (1/2)at^2
d = 0 * 0.35 s + (1/2) * 20 m/s^2 * (0.35 s)^2
d = (1/2) * 20 m/s^2 * 0.1225 s^2
d = 1.225 m
Therefore, the shuffleboard disc travels a distance of approximately 1.225 meters after the push stops.