Question

Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a distance of 14.7 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0376. Initially, how fast was the skier going?

Answer 1

Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a distance of $14.7\mathrm{m}$ before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0376. Initially, how fast was the skier going?

The law of conservation of energy:

$$\Delta E + W = 0$$

where $\Delta E$ – change of body's energy, $W$ – work of all forces

Work can be expressed by the following equation:

$$W = F d \cos \theta$$

where $F$ is the force, $d$ is the displacement, and the angle $\theta$ is defined as the angle between the force and the displacement vector.

Therefore, for friction force work equals:

$$W = F _ {f r} * d * \cos 180 = - F _ {f r} d$$

Friction force equals:

$$F _ {f r} = \mu m g$$

Change of body's energy equals $\frac{mv^2}{2}$, therefore:

$$\frac {m v ^ {2}}{2} = \mu m g d$$ $$v = \sqrt {2 \mu g d} = 2.33 \frac {m}{s}$$

Answer: $2.33\frac{m}{s}$