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(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?

Short Answer

Expert verified

The skier’s velocity decrease by \(5.77\;{\rm{m/s}}\).

Step by step solution

01

Momentum and impulse relation

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.

02

Given data

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.

03

Calculation of the change in velocity

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 \(5.77\;{\rm{m/s}}\).

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Most popular questions from this chapter

A golf ball of mass 0.045 kg is hit off the tee at a speed of 38 m/s. The golf club was in contact with the ball for \({\bf{3}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\;{\bf{s}}\). Find

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FIGURE 7-34 Problem 24.

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