# Sam, whose mass is 68 kg , straps on his

Sam, whose mass is 68 kg , straps on his skis and starts down a 58 mm -high, 20∘ frictionless slope. A strong headwind exerts a horizontal force of 200 NN on him as he skis. Use work and energy to find Sam's speed at the bottom.
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Neunassauk8
Given that, mass of the Sam is m=68 kg
Angle of the slope, $\theta ={20}^{\circ }$
Height of the slope, h=58 mm
Length of the slope is given by
$d=\frac{h}{\mathrm{sin}\theta }$

From work energy theorem, the net work done on the Sam will be equal to the change in kinetic energy.
${W}_{g}-{W}_{w\in d}=\mathrm{△}K$
The force acting on the skier is the weight of the skier downwards along the slope and the component of force along the slope. Thus the above equation is written as
$\left[mg\mathrm{sin}\theta -{F}_{w}\mathrm{cos}\theta \right]d=\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}$
$\left[mg\mathrm{sin}\theta -{F}_{w}\mathrm{cos}\theta \right]d=\frac{1}{2}m{v}_{f}^{2}-0$
${v}_{f}^{2}=\frac{2\left[mg\mathrm{sin}\theta -{F}_{w}\mathrm{cos}\theta \right]d}{m}$
${v}_{f}=\sqrt{\frac{2\left[mg\mathrm{sin}\theta -{F}_{w}\mathrm{cos}\theta \right]d}{m}}$
$=\sqrt{\frac{2\left(68\right)\left(9.8\right)\left({\mathrm{sin}20}^{\circ }\right)-\left(200\right){\mathrm{cos}20}^{\circ }}{0.169}\left\{68\right\}}$

The Sam speed at the bottom is 0.445 m/s