Question

Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the surface of the slope. The combined mass of skis and Kieran is 50 kg (the fuel mass is negligible). Kieran's speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow?

Answer 1

Answer:0.46

Explanation:

Given

Initial height $h=50 m$

inclination $\theta =10^{\circ}$

Thrust$=280 N$

combined mass of kieran and skis $m=50 kg$

Speed at the bottom $v=40 m/s$

From Work Energy Theorem

Work done by all the force is equal to change in kinetic Energy

$W_{gravity}+W_{friction}+W_{thrust}=\frac{1}{2}mv^2-0$------------1

distance traveled along the slope $x=\frac{50}{\sin 10}=287.93 m$

$W_{gravity}=mgh=50\times 9.8\times 50=24500 J$

$W_{thrust}=F\times x=280\times 287.93=80,620.4 J$

$W_{friction}=-\mu mg\cos 10$

substitute in 1

$24,500+80,620.4+W_{friction}=\frac{1}{2}\times 50\times 1600$

$W_{friction}=40,000-24,500-80,620.4$

$-\mu \cdot 50\times 9.8\times 287.93=-65,120.4$

$\mu =\frac{65,120.4}{141,085.7}=0.46$