A hockey puck is hit on a frozen lake and starts moving with a speed of 12.3 m/s. Five seconds later, its speed is 6.60 m/s.
(a) What is its average acceleration?
1 m/s2
(b) What is the average value of the coefficient of kinetic friction between puck and ice?
2
(c) How far does the puck travel during the 5.00 s interval?
3 m
(a) Acceleration is the change in speed (V2 - V1) divided by the time interval of the change (T = 5 s). In this case, it is negative.
(b)(Average friction force)*time = momentum change
M*g*mu*T = M(V1 - V2)
mu = (V1 -V2)/(g*T)
(c) Average speed * Time = (V1 + V2)*T/2
To find the average acceleration, you need to calculate the change in velocity and divide it by the change in time.
(a) Average acceleration = (final velocity - initial velocity) / (final time - initial time)
= (6.60 m/s - 12.3 m/s) / (5 s - 0 s)
= -5.7 m/s / 5 s
= -1.14 m/s²
Therefore, the average acceleration of the puck is -1.14 m/s².
To find the average value of the coefficient of kinetic friction, you can use the formula for acceleration due to friction:
(b) Acceleration due to friction = μ * g
where μ is the coefficient of kinetic friction and g is the acceleration due to gravity.
Since the puck is moving horizontally, the only force acting on it is the force of friction. Therefore, the acceleration due to friction is equal to the average acceleration of the puck.
Hence, -1.14 m/s² = μ * g
To find the average value of the coefficient of kinetic friction, you need the value of acceleration due to gravity.
(c) The distance traveled by the puck during the 5-second interval can be found using the formula:
Distance = Initial velocity * time + (1/2) * acceleration * time²
Since the acceleration is constant, you can use the average acceleration value obtained in part (a).
Distance = (12.3 m/s) * 5 s + (1/2) * (-1.14 m/s²) * (5 s)²
= 61.5 m - 14.25 m
= 47.25 m
Therefore, the puck travels a distance of 47.25 meters during the 5.00-second interval.