Answer to Question #162687 in Mechanics | Relativity for Adunola

Question

 A sleigh ride takes place on an even, inclined, straight toboggan run with a gradient of 3: 100. The sled with its crew has a mass of m = 80 kg and a coefficient of sliding friction of u = 0.025. After a short push, the static friction is overcome and an evenly accelerated movement begins.

 1) Calculate the downhill force, the frictional force and the resulting acceleration a, of the slide.

 (3 P.) 2) After a time and a covered distance of s, the slide reaches a speed of Vmax = 40 km / h. Calculate, and s ,.

 (3 P.) 3) The sledge is braked so that it continues to toboggan for 667 m. Calculate the required time t,. The slide is then braked evenly for 30 s until it comes to a standstill. It covers a distance of sz with a constant braking deceleration ag. Calculate są and az as well as the total length and total time of the toboggan run.

 (9 P.) 4) Make sketches with inscription of the speed-time-diagram and the distance-time-diagram. Choose suitable scales yourself!

 (4 p.)

Answer 1

Given,

Inclination $\tan \theta = \frac{3}{10}$

$\sin \theta = \frac{3}{\sqrt{109}}$

$\cos\theta = \frac{10}{\sqrt{109}}$

Mass of the sled crew (M)=80kg

Coefficient of sliding friction $(\mu) =0.025$

So, downward force $F=mg\sin\theta -\mu mg \cos\theta$

$=80\times 10\frac{3}{\sqrt{109}}-0.025\times 10\times 80\times\frac{10}{\sqrt{109}}$

$=80\frac{(3-0.25)}{\sqrt{109}}$

$=\frac{80\times 2.75}{\sqrt{109}}$ N

Now,

$a=\frac{2.75}{\sqrt{109}}m/s^2$

$\Rightarrow s=\frac{v^2}{2a}$

$\Rightarrow s = \frac{40^2}{2\times \sqrt{109}}m$

$s=\frac{800}{\sqrt{109}}m$

Now it is given s = 667

$t=\sqrt{\frac{2s}{a}}$

$t=\sqrt{\frac{2\times667\times\sqrt{109}}{2.75}}s$