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# A hockey puck is hit on a frozen lake and starts moving with a speed

A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0m/s. Five seconds later, its speed is 6.00m/s. What is its average acceleration? What is the average value of the coefficient of kinetic friction between puck and ice? How far does the puck travel during the 5.00s interval?

Equation for speed as a function of time for constant acceleration is:

v = u + at where v is the final speed, u is the initial speed, a is the acceleration, and t is the time elapsed. This gives a = (v - u)/t

In this case, v = 6 m/s, u = 12 m/s, t = 5s.
Thus a = (6 - 12) / 5 = -1.2 m/s^2
Average acceleration = - 1.2 m/s^2
(The negative sign means it is decelerating).

From laws of motion we have F = ma, where F is the force, m is the mass, and a is the acceleration. So we have:

F = -1.2m with m being the mass of the puck. The force is negative because it acts in a direction opposite to the motion of the puck.

The magnitude of the force is the force of friction. The friction force Q is given as:
Q = cmg, where c is coeff of friction, m is the mass of the puck, and g is the gravitational acceleration (9.8 m/s^2). Thus we have:

1.2m = cmg

which gives c = 1.2/g = 1.2/9.8 = 0.1224
Thus, coeff of kinetic friction is 0.1224.

Distance traveled d is given by:

d = ut + 0.5at^2
Again, u = 12 m/s, a = -1.2 m/s^2, t = 5s. This gives d = 12x5 - 0.5x1.2x25 = 45 m
Distance traveled by puck = 45 m

Customer: replied 7 years ago.

That was quick! i now know where i went wrong with this problem.

thankyou.

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