Answer:
a)The initial velocity of the puck is 20 ft/s.
b)The coefficient of friction is 0.062.
Explanation:
Hi there!
a)For this problem let's use the equations of position and velocity of an object moving in a straight line with constant acceleration:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the puck after a time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
v = velocity of the puck at a time t.
Let's place the origin of the frame of reference at the point where the puck is hit so that x0 = 0.
We know that at t = 10 s the velocity of the puck is zero (v = 0) and its position is 100 ft (x = 100 ft):
100 ft = v0 · 10 s + 1/2 · a · (10 s)²
0 = v0 + a · 10 s
We have a system of two equations with two unknowns, so, we can solve the system.
Solving for v0 in the second equation:
0 = v0 + a · 10 s
v0 = -a · 10 s
Replacing v0 in the first equation:
100 ft = (-a · 10 s) · 10 s + 1/2 · a · (10 s)²
100 ft = -50 s² · a
100 ft / -50 s² = a
a = -2.0 ft/s²
Then the initial velocity of the puck will be:
v0 = -a · 10 s
v0 = -(-2.0 ft/s²) · 10 s
v0 = 20 ft/s
The initial velocity of the puck is 20 ft/s.
b) The friction force is calculated as follows:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of friction.
Since the only vertical forces acting on the puck are the weight of the puck and the normal force and since the puck is not being accelerated in the vertical direction, then, the normal force is equal to the weight of the puck. The weight (W) is calculated as follows:
W = m · g
Where "m" is the mass of the puck and "g" is the acceleration due to gravity (32.2 ft/s²).
Then the friction force can be calculated as follows:
Fr = m · g · μ
Since the acceleration of the puck is provided only by the friction force, then, due to Newton's second law:
Fr = m · a
Where "m" is the mass of the puck and "a" its acceleration. Then:
Fr = m · g · μ
Fr = m · a
m · g · μ = m · a
μ = a/g
μ = 2.0 ft/s² / 32.2 ft/s²
μ = 0.062
The coefficient of friction is 0.062.
