(I) A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity?

According to Newton’s second law, the impulse on an object is equal to the change in momentum. The impulse is the product of force and time.

The mass of the skier is \(m = 65\;{\rm{kg}}\).

The constant force on the skier is \(F = 25\;{\rm{N}}\).

The time duration of the force is \(\Delta t = 15\;{\rm{s}}\).

Let \(\Delta v\) be the change in the velocity of the skier.

The impulse of the force acting on the skier is\(F\Delta t\)and the magnitude of the change in momentum of the skier is\(m\Delta v\).

The impulse is equal to a change in momentum. Then,

\(\begin{array}{c}m\Delta v = F\Delta t\\\left( {65\;{\rm{kg}}} \right) \times \Delta v = \left( {25\;{\rm{N}}} \right) \times \left( {15\;{\rm{s}}} \right)\\\left( {65\;{\rm{kg}}} \right) \times \Delta v = \left( {25\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {15\;{\rm{s}}} \right)\\\Delta v = 5.77\;{\rm{m/s}}\end{array}\)

The frictional force acts in the opposite direction of motion. Therefore, the velocity decreases for this impulse.

Hence, the skier’s velocity decreases by \(5.77\;{\rm{m/s}}\).