What force is required to keep a 75 kg man skating at 3.0 m/s if the coefficient of kineticfriction for ice on steel is 0.050?
Question
What force is required to keep a 75 kg man skating at 3.0 m/s if the coefficient of kineticfriction for ice on steel is 0.050?
Solution
To solve this problem, we need to use the formula for the force of friction, which is:
F_friction = μk * F_normal
where:
- F_friction is the force of friction,
- μk is the coefficient of kinetic friction,
- F_normal is the normal force.
Step 1: Calculate the normal force. For an object on a flat surface, the normal force is equal to the weight of the object. So, we have:
F_normal = mass * g F_normal = 75 kg * 9.8 m/s² F_normal = 735 N
Step 2: Substitute the normal force and the coefficient of kinetic friction into the formula for the force of friction:
F_friction = μk * F_normal F_friction = 0.050 * 735 N F_friction = 36.75 N
So, the force required to keep the man skating at a constant velocity is 36.75 N.
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