A 84.5 kg skater moving initially at 2.40 m/s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice.What force does friction exert on the skater?
To find the force exerted by friction on the skater, we can use Newton's second law of motion, which states that the net force (F) acting on an object is equal to its mass (m) multiplied by its acceleration (a):
F = ma
In this case, the skater comes to rest uniformly, so the acceleration is constant and equal to 0. Therefore, the net force acting on the skater is also 0. However, the frictional force is the force that opposes the motion of the skater and eventually brings it to rest. Therefore, the magnitude of the frictional force is equal to the force needed to bring the skater to rest, which is equal to the force of kinetic friction.
The force of kinetic friction (fK) can be calculated using the equation:
fK = μk * N
where μk is the coefficient of kinetic friction and N is the normal force.
Since the skater is on rough horizontal ice, the normal force is equal to the skater's weight, which can be calculated using the equation:
N = mg
where m is the mass of the skater and g is the acceleration due to gravity.
Given:
- Mass of the skater (m) = 84.5 kg
- Initial velocity (u) = 2.40 m/s
- Final velocity (v) = 0 m/s
- Time (t) = 3.52 s
To find the coefficient of kinetic friction (μk), we can use the equation:
v = u + at
Since the final velocity (v) is 0 and the initial velocity (u) is given as 2.40 m/s, we have:
0 = 2.40 + a * 3.52
Solving for a:
a = -2.40 / 3.52 = - 0.682 m/s^2
Now we can use this acceleration value to calculate the coefficient of kinetic friction (μk):
a = μk * g
-0.682 = μk * 9.8
Solving for μk:
μk = -0.682 / 9.8 = -0.0696
Note: The coefficient of kinetic friction cannot be negative, so we take the absolute value:
μk = 0.0696
Now that we have the coefficient of kinetic friction, we can calculate the force of kinetic friction (fK):
fK = μk * N
N = mg
N = 84.5 kg * 9.8 m/s^2 = 828.1 N
fK = 0.0696 * 828.1 N = 57.6 N
Therefore, the force of friction exerted on the skater is 57.6 N.