A hockey puck is hit on a frozen lake and
starts moving with a speed of 12.1 m/s. Four
seconds later, its speed is 6.7 m/s.
What is its average acceleration? The ac-
celeration of gravity is 9.8 m/s2 .
Answer in units of m/s2.
What is the average value of the coefficient of
kinetic friction between puck and ice?
How far does the puck travel during this 4 s
interval?
Answer in units of m.
a = change in v /change in time
= (6.7 - 12.1)/4
= -1.35 m/s^2
F = m a = - mu m g
so
mu = 1.35/9.8 = .138
average speed = (12.1+6.7)/2 = 9.4
9.4*4 = 37.6 meters
Thank you! That makes much more sense!
To find the average acceleration of the hockey puck, we can use the formula:
average acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity is 12.1 m/s, the final velocity is 6.7 m/s, and the time is 4 seconds. Plugging these values into the formula, we get:
average acceleration = (6.7 m/s - 12.1 m/s) / 4 s
Simplifying, we have:
average acceleration = (-5.4 m/s) / 4 s
Therefore, the average acceleration of the hockey puck is -1.35 m/s^2.
To find the average value of the coefficient of kinetic friction between the puck and the ice, we need to consider the relationship between acceleration, friction, and the weight of the puck. The equation for the net force acting on an object on a horizontal surface is:
net force = mass * acceleration
Since the puck is not accelerating vertically and there is no vertical force, the weight of the puck can balance the vertical forces:
weight = mass * gravitational acceleration
Now, we can include the friction force:
friction force = coefficient of kinetic friction * normal force
The normal force is equal to the weight (since there is no vertical acceleration), so we can rewrite the friction force equation as:
friction force = coefficient of kinetic friction * weight
Substituting the equation for net force and the equation for friction force, we get:
mass * acceleration = coefficient of kinetic friction * mass * gravitational acceleration
Simplifying, we find:
acceleration = coefficient of kinetic friction * gravitational acceleration
Rearranging the equation to solve for the coefficient of kinetic friction, we have:
coefficient of kinetic friction = acceleration / gravitational acceleration
Plugging in the values for acceleration (which we found previously as -1.35 m/s^2) and gravitational acceleration (which is given as 9.8 m/s^2), we get:
coefficient of kinetic friction = (-1.35 m/s^2) / 9.8 m/s^2
Simplifying, we find:
coefficient of kinetic friction ≈ -0.138
The negative sign indicates that the coefficient of kinetic friction is in the opposite direction to the motion of the puck.
To find the distance the puck travels during the 4-second interval, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is 12.1 m/s, the time is 4 seconds, and the acceleration is -1.35 m/s^2 (which we found earlier). Plugging in these values, we get:
distance = (12.1 m/s) * 4 s + (1/2) * (-1.35 m/s^2) * (4 s)^2
Simplifying, we have:
distance = 48.4 m - 4.32 m
Therefore, the puck travels approximately 44.08 meters during the 4-second interval.