What distance will a sliding hockey puck with a kinetic energy of 500J go if it’s experiencing a normal force of 20N on ice with a friction constant of 0.1?

The distance of this sliding hockey puck will be 250 meters. This is because of the law of conservation of energy, which states that the kinetic energy is equal to the normal force times the coefficient of friction times the distance traveled in the end.

Expert Answers

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To answer this, we will look at the kinematic equations and explore each of the different terms/concepts presented in the question.

For starters, kinetic energy is equal to 1/2 x mass x velocity squared, or 1/2mv^2. Additionally, the coefficient of friction tells you how much force will be applied against the direction of motion, and it is a factor of the normal force.

One of the kinematic equations is (initial velocity squared) = (final velocity squared) + (2 x acceleration x distance). This can be reconfigured to show that initial kinetic energy is equal to the mass times the acceleration times the distance traveled, for an object slowing down to a stop (since the final velocity will be zero).

Therefore, 500J = 20N x 0.1 x distance. This simplifies to 500J = 2N x distance in meters, or 250 meters.

Because the mass of the object remains constant, it is not necessary to solve for it within the kinematic equation, even though it is present. In the end, the equation simplifies rather easily to show that the normal force times the coefficient of friction times the distance is equal to the initial kinetic energy. This is a direct result of the low of conservation of energy.

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