A hockey puck is hit on a frozen lake and starts moving with a speed of 10.6 m/s. Five seconds later, its speed is 6.10 m/s.
(a) What is its average acceleration?
m/s2
(b) What is the average value of the coefficient of kinetic friction between puck and ice?
(c) How far does the puck travel during this 5.00 s interval?
m
To solve this problem, we can use the equations of motion for uniformly accelerated motion:
(1) v = u + at
(2) s = ut + (1/2)at^2
(3) v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance
Let's solve the problem step-by-step:
(a) To find the average acceleration, we can use formula (1). Rearranging the equation to solve for acceleration:
a = (v - u) / t
where:
v = 6.10 m/s (final velocity)
u = 10.6 m/s (initial velocity)
t = 5.00 s (time)
Substituting the values:
a = (6.10 - 10.6) / 5.00
a = (-4.50) / 5.00
a = -0.90 m/s^2
Therefore, the average acceleration is -0.90 m/s^2 (negative sign indicates a deceleration).
(b) To find the average value of the coefficient of kinetic friction, we can use the equation:
a = -μg
where:
a = acceleration
μ = coefficient of kinetic friction
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values:
-0.90 = -μ * 9.8
Solving for μ:
μ = -0.90 / 9.8
μ ≈ 0.0918
Therefore, the average value of the coefficient of kinetic friction between the puck and ice is approximately 0.0918.
(c) To find the distance traveled by the puck during the 5.00 s interval, we can use formula (2):
s = ut + (1/2)at^2
where:
u = 10.6 m/s (initial velocity)
t = 5.00 s (time)
a = -0.90 m/s^2 (average acceleration)
Substituting the values:
s = (10.6 * 5.00) + (1/2)(-0.90)(5.00^2)
s = 53.0 + (1/2)(-0.90)(25.0)
s = 53.0 - (0.45)(25.0)
s = 53.0 - 11.25
s ≈ 41.75 m
Therefore, the puck travels approximately 41.75 meters during the 5.00 s interval.
To find the answers to these questions, we will need to use the equations of motion and the relationship between acceleration, velocity, and displacement. Let's break down each question step by step.
(a) What is its average acceleration?
Average acceleration is calculated by finding the change in velocity divided by the time taken. We are given the initial velocity (u = 10.6 m/s) and the final velocity (v = 6.10 m/s) and the time taken (t = 5 seconds).
Using the equation for average acceleration (a), we can plug in the values:
a = (v - u) / t
a = (6.10 m/s - 10.6 m/s) / 5 s
Calculating this expression gives:
a = (-4.5 m/s) / 5 s
a = -0.9 m/s²
Therefore, the puck's average acceleration is -0.9 m/s² (note the negative sign indicates it is decelerating).
(b) What is the average value of the coefficient of kinetic friction between the puck and ice?
The average value of the coefficient of kinetic friction can be calculated using the formula:
µk = a / g
Where a is the acceleration and g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the value of a from part (a):
µk = -0.9 m/s² / 9.8 m/s²
Calculating this expression gives:
µk = -0.0918
Therefore, the average value of the coefficient of kinetic friction between the puck and ice is -0.0918.
(c) How far does the puck travel during this 5.00 s interval?
To calculate the distance traveled by the puck, we can use the equation:
s = ut + (1/2)at²
Where s represents the displacement, u is the initial velocity, t is the time, and a is the acceleration.
Plugging in the values:
s = (10.6 m/s)(5 s) + (1/2)(-0.9 m/s²)(5 s)²
Calculating this expression gives:
s = 53 m + (-11.25 m)
s = 41.75 m
Therefore, the puck travels a distance of 41.75 meters during the 5-second interval.
a. a = (Vf - Vo) / t,
a = (6.1 - 10.6) / 5 = -0.9m/s^2.
c. d = (Vf^2 - Vo^2) / 2a,
d = ((6.1)^2 - (10.6)^2) / -1.8 = 41.8m.