Holding on to a towrope moving parallel to a frictionless ski slope, askier is pulled up the slope, which is at an angle of$\mathbf{8}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}$with the horizontal. What is the magnitude of the force on the skier from the rope when (a) the magnitude vof the skier’s velocity is constant atlocalid="1657161930942" $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and (b) as vincreases at a rate oflocalid="1657161837037" $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

• The mass of the skier is,$m=50kg.$
• The inclination angle with the horizontal is,$\theta =8.0^0.$
• The initial velocity of the skier is,$v=2.0\mathrm{m}/\mathrm{s}.$
• The acceleration of the skier is,role="math" localid="1657163136849" $a=0.10\mathrm{m}/{\mathrm{s}}^2.$

A free body diagram has to be drawn. By using Newton’s second law and vector resolution concept, the force on the skier at constant velocity and with varying velocity can be found.

Formula:

$\overrightarrow{F_{net}}=m\overrightarrow{a}$

By applying Newton’s second law along x axis as,

$\overrightarrow{F_{net}}=m\overrightarrow{a}$

The skier is moving with constant velocity hence it has zero acceleration. Consider the sign convention according to the direction of the skier,

role="math" localid="1657164170548" $$\begin{gathered} F_{rope}-mg\sin \theta =0 \\ {F}_{rope}=mg\sin \theta \\ =50kg\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin 8.0^0 \\ =68\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the rope force is 68 N.

The skier is going up with the rate of acceleration, hence according to the Newton’s second law,

$$\begin{gathered} F_{rope}-mg\sin \theta =ma \\ {F}_{rope}=ma+mg\sin \theta \\ =50kg\times 0.10\mathrm{m}/{\mathrm{s}}^2+50\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin (8.0^0) \\ =73\mathrm{N} \end{gathered}$$

Hence, the magnitude of the rope force, in this case, will be73 N .