Answer:
The required net force has a magnitude of 7200 N
Explanation:
Use Newton's 2nd Law to obtain the answer:
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Answer:
a. The acceleration of the hockey puck is -0.125 m/s².
b. The kinetic frictional force needed is 0.0625 N
c. The coefficient of friction between the ice and puck, is approximately 0.012755
d. The acceleration is -0.125 m/s²
The frictional force is 0.125 N
The coefficient of friction is approximately 0.012755
Explanation:
a. The given parameters are;
The mass of the hockey puck, m = 0.5 kg
The starting velocity of the hockey puck, u = 5 m/s
The distance the puck slides and slows for, s = 100 meters
The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration
The velocity of the hockey puck after motion, v = 0 m/s
The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;
v² = u² + 2·a·s
Therefore, by substituting the known values, we have;
0² = 5² + 2 × a × 100
-(5²) = 2 × a × 100
-25 = 200·a
a = -25/200 = -0.125
The constant acceleration of the hockey puck, a = -0.125 m/s².
b. The kinetic frictional force, , required is given by the formula, F = m × a,
From which we have;
= 0.5 × 0.125 = 0.0625 N
The kinetic frictional force required, = 0.0625 N
c. The coefficient of friction between the ice and puck, , is given from the equation for the kinetic friction force as follows;
The coefficient of friction between the ice and puck, ≈ 0.012755
d. When the mass of the hockey puck is 1 kg, we have;
Given that the coefficient of friction is constant, we have;
The frictional force
The acceleration, a = /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a
The frictional force as calculated here,
The coefficient of friction ≈ 0.012755 is constant