Question

A 65.3-kg skier coasts up a snow-covered hill that makes an angle of 35.0 ° with the horizontal. The initial speed of the skier is 9.64 m/s. After coasting a distance of 2.37 m up the slope, the speed of the skier is 4.00 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Answer 1

A 65.3-kg skier coasts up a snow-covered hill that makes an angle of $35.0{}^{\circ}$ with the horizontal. The initial speed of the skier is $9.64\mathrm{m/s}$. After coasting a distance of $2.37\mathrm{m}$ up the slope, the speed of the skier is $4.00\mathrm{m/s}$. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Solution

(a)

Use energy equation:

Let U (potential energy) be 0 at the bottom of the hill and be of the form $m * g * h$ elsewhere, K is kinetic energy.

$$\mathrm{So} \left(K + U\right)_{1} + W_{friction} = \left(K + U\right)_{2}$$

or

$$W_{friction} = K_{2} + U_{2} - K_{1} - U_{1} = \frac{1}{2} * m * v_{2}^{2} + m * g * d * \sin(\theta) - \frac{1}{2} * m * v_{1}^{2}.$$ $$W_{friction} = \frac{1}{2} * 65.3 * 4.00^{2} + 65.3 * 9.80 * 2.37 * \sin 35.0{}^{\circ} - \frac{1}{2} * 65.3 * 9.64^{2} = -1642 \, \text{J}.$$

(Negative because it removes energy from the system)

(b) Now $W_{friction} = F * d$, so

$$F = \frac{W_{friction}}{d} = \frac{1642}{2.37} = 693 \, \text{N}.$$

Answer: (a) $-1642 \, \text{J}$; (b) $693 \, \text{N}$.