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 $g = 10\\ m/s^{2}$.

a) draw a free - body diagram for the skier, and find the weight and normal force acting on the skier.
b) find the acceleration of the skier down the slope. (bonus, what would the acceleration be if the skier has three times the mass?)

1. repo. a fancy car (mass = 1500 kg) is connected via a chain to a tow truck and is towed up a hill at a constant velocity. the slope of the hill is 20°. friction is negligible.

a) draw a free - body diagram for the car, and calculate the tension in the chain.

Explanation:

Step1: Calculate skier's weight

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

Step2: Analyze forces perpendicular to the slope for normal force

In the direction perpendicular to the slope, the normal force $N$ balances the component of the weight perpendicular to the slope. Using $N = mg\cos\theta$, with $\theta = 22^{\circ}$, $m = 50$ kg and $g = 10$ m/s², we have $N=50\times10\times\cos(22^{\circ})\approx50\times10\times0.9272 = 463.6$ N.

Step3: Analyze forces parallel to the slope for acceleration

In the direction parallel to the slope, according to Newton's second - law $\sum F=ma$. The net force parallel to the slope is $F_{net}=mg\sin\theta$. So $mg\sin\theta=ma$, and $a = g\sin\theta$. Substituting $\theta = 22^{\circ}$ and $g = 10$ m/s², we get $a = 10\times\sin(22^{\circ})\approx10\times0.3746=3.746$ m/s². The acceleration is independent of mass, so if the mass is three - times, the acceleration remains $a = 10\times\sin(22^{\circ})\approx3.746$ m/s².

Step4: Analyze forces for the car

For the car moving at a constant velocity, the net force is zero. In the direction parallel to the slope, the tension $T$ in the chain balances the component of the car's weight parallel to the slope. Using $T = mg\sin\theta$, with $m = 1500$ kg, $\theta = 20^{\circ}$ and $g = 10$ m/s², we have $T=1500\times10\times\sin(20^{\circ})\approx1500\times10\times0.3420 = 5130$ N.

Answer:

a) The weight of the skier is 500 N and the normal force is approximately 463.6 N.
b) The acceleration of the skier down the slope is approximately 3.746 m/s², and it remains the same if the skier has three - times the mass.
2a) The tension in the chain is 5130 N.