Question

a 55-kg person is skating on the rideau canal. she reaches a maximum speed and then glides to a stop in 100 m. the coefficient of friction between her skates and the ice is 0.010. what is the work done by the kinetic friction?

Answer 1

The work done by kinetic friction on a 55-kg person who stops after gliding 100 m on ice, with a friction coefficient of 0.010, is 539 Joules.

Step-by-step explanation:

To determine the work done by kinetic friction on a 55-kg person who glides to a stop on the ice, we can use the formula W = f × d, where W is the work done by friction, f is the frictional force, and d is the distance over which the force is applied. The frictional force can be found through the equation f = μ × m × g, where μ is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity (9.8 m/s2). Given that μ = 0.010 and the person glides 100 m, the frictional force f is:

f = 0.010 × 55 kg × 9.8 m/s2 = 5.39 N

Thus, the work done by kinetic friction is:

W = 5.39 N × 100 m = 539 J

The work done is 539 Joules (J), which is negative since friction does work against the motion of the skater.