Question

7. A 61-kg skier, coasting down a hill that is at an angle of ${23}^{\circ }$ to the horizontal, experiences a force of kinetic friction of magnitude $72\mathrm{N}$. The skier's speed is $3.5\mathrm{m}/\mathrm{s}$ near the top of the slope. Determine the speed after the skier has travelled $62\mathrm{m}$ downhill. Air resistance is negligible.

Answer 1

To find the speed of the skier after traveling 62m downhill, we need to consider the conservation of energy. The initial energy of the skier at the top of the hill is entirely potential energy, which is converted to kinetic energy as the skier moves downhill.

The initial potential energy of the skier is given by:

$$PE=mgh$$
$$PE=(61\mathrm{kg})(9.81\mathrm{m}/{\mathrm{s}}^2)(62\mathrm{m})$$
$$PE=37858.02\mathrm{J}$$

At the bottom of the hill, all of the potential energy has been converted to kinetic energy:

$$KE=\frac{1}{2}m{v}^2$$
$$37858.02=\frac{1}{2}(61)v^2$$
$$v=\sqrt{\frac{2(37858.02)}{61}}$$
$$v\approx 26.37\mathrm{m}/\mathrm{s}$$

Therefore, the speed of the skier after traveling 62m downhill is approximately $26.37\mathrm{m}/\mathrm{s}$.