A hockey puck is hit on a frozen lake and starts moving with a speed of 14.0 m/s. Five seconds later, its speed is 5.50 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 the 5.00 s interval?
m
for a) i got -1.70m/s^2
b) i need help
c) i got 48.8m
so can anyone help me with What is the average value of the coefficient of kinetic friction between puck and ice?
see other post
To find the answers to these questions, we need to use the kinematic equations and the concept of average acceleration.
(a) Let's start by finding the average acceleration of the hockey puck. The average acceleration can be calculated using the formula:
average acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity of the puck is 14.0 m/s, the final velocity is 5.50 m/s, and the time is 5 seconds. Plugging these values into the formula, we get:
average acceleration = (5.50 m/s - 14.0 m/s) / 5 s
average acceleration = -8.5 m/s / 5 s
average acceleration = -1.7 m/s^2
Therefore, the average acceleration of the hockey puck is -1.7 m/s^2.
(b) To find the average value of the coefficient of kinetic friction between the puck and ice, we need to use an equation that involves acceleration and the coefficient of kinetic friction. The equation we can use is:
acceleration = -μ * g
Where μ is the coefficient of kinetic friction and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the acceleration is -1.7 m/s^2. Plugging this value into the equation, we can solve for μ:
-1.7 m/s^2 = -μ * 9.8 m/s^2
Dividing both sides of the equation by -9.8 m/s^2:
μ = (-1.7 m/s^2) / (-9.8 m/s^2)
μ = 0.173
Therefore, the average value of the coefficient of kinetic friction between the puck and ice is 0.173.
(c) To find the distance the puck travels during the 5.00 s interval, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is 14.0 m/s, the time is 5.00 s, and the acceleration is -1.70 m/s^2. Plugging these values into the equation, we get:
distance = (14.0 m/s) * (5.0 s) + (1/2) * (-1.70 m/s^2) * (5.0 s)^2
distance = 70 m - (0.85 m/s^2) * 25 s^2
distance = 70 m - 21.25 m
distance = 48.75 m
Therefore, the puck travels a distance of 48.75 meters during the 5.00 s interval.