Question

worksheet 2.3: force vectors, part 2
$v=\frac{d}{t}$ $a = \frac{delta v}{t}$ $d=\frac{1}{2}at^{2}$ $sigma f=ma$

1. black diamond. consider a skier going down a 22° slope (this is considered quite steep). the skiers mass is 50 kg. the friction between the snow and skis is negligible. you may use g = 10 m/s².

a) draw a free - body diagram for the skier, and find the weight and normal force acting on the skier.

Explanation:

Step1: Calculate the weight of the skier

The weight formula is $W = mg$. Given $m = 50$ kg and $g=10$ m/s², we have $W=50\times10 = 500$ N.

Step2: Analyze forces perpendicular to the slope

The normal force $N$ on the skier perpendicular to the slope can be found by considering the equilibrium of forces perpendicular to the slope. The component of the weight perpendicular to the slope is $mg\cos\theta$. Since there is no acceleration in the direction perpendicular to the slope, $N = mg\cos\theta$. Substituting $m = 50$ kg, $g = 10$ m/s² and $\theta=22^{\circ}$, we get $N=50\times10\times\cos(22^{\circ})\approx50\times10\times0.9272 = 463.6$ N.

Answer:

Weight: 500 N; Normal force: approximately 463.6 N.
(Note: To draw the free - body diagram, draw a dot to represent the skier. Draw a vector straight down to represent the weight $W = 500$ N. Draw a vector perpendicular to the slope pointing away from the slope to represent the normal force $N\approx463.6$ N.)