A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. Five seconds later, its speed is 6.00 m/s. (a) What is its average acceleration? (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?

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Chris Cheggies

Beuehe

Darkie

To find the average acceleration, we can use the formula:

Acceleration (a) = (Final Velocity - Initial Velocity) / Time

a = (6.00 m/s - 12.0 m/s) / 5 s
a = (-6.00 m/s) / 5 s
a = -1.20 m/s²

Therefore, the average acceleration of the hockey puck is -1.20 m/s². The negative sign indicates that the puck is decelerating.

To find the average value of the coefficient of kinetic friction between the puck and the ice, we use the equation:

Frictional Force (Ff) = Coefficient of Friction (μ) * Normal Force (Fn)

Since the ice is horizontal, the normal force (Fn) is equal to the weight of the puck:

Fn = mass * gravity

We need the mass of the puck to calculate Fn. If that information is not provided, we won't be able to determine the coefficient of kinetic friction.

Lastly, to find the distance traveled by the puck during the 5.00-second interval, we can use the formula:

Distance (d) = Initial Velocity * Time + 0.5 * Acceleration * Time²

d = (12.0 m/s) * 5 s + 0.5 * (-1.20 m/s²) * (5 s)²

Simplifying the equation gives:

d = 60 m - 7.5 m
d = 52.5 m

Therefore, the puck travels a distance of 52.5 meters during the 5.00-second interval.