A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake at a constant velocity.

Question

A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake at a constant velocity. If the coefficient of friction of ice on ice is 0.03, then what is the magnitude of force applied by the person on the block of ice?

Answer 1

Some key words to notice in the question here are "constant velocity", "friction", and "force applied". If the block is moving at a constant velocity, then the block has an acceleration of 0. If the acceleration is 0, that means that the net force is also equal to 0. We know that we're applying a force to push the block, so in order for it to have a net force of 0, something has to be equal and opposite to our applied force. This is going to be the force of friction.

We find the force of friction using F=μN. Here we have a flat surface, so the normal force is equal to the weight of the block, mass x gravitational acceleration. Therefore, we get F=μmg for force of friction. We said earlier that the force of friction is equal to the force that we are applying, so we can use this equation to simply find the answer.

F = μmg = (0.03)(45)(9.81) = 13.24 N