Question

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of${\mathbf{12}}^{\mathbf{°}}$with the horizontal. The rope moves parallel to the slope with a constant speed of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. The force of the rope does $\mathbf{900}\mathbf{}\mathbf{J}$of work on the skier as the skier moves a distance of $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$up the incline. (a) If the rope moved with a constant speed of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, how much work would the force of the rope do on the skier as the skier moved a distance of $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b)$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and (c)$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$?

Short answer

1. The work done by the force of the rope on the skier at speed $2.0\mathrm{m}/\mathrm{s}$is same that when speed is 1.0 m/s.
2. The rate of doing work when rope moves with speed of$1.0\mathrm{m}/\mathrm{s}\mathrm{is}112.5\mathrm{W}$
3. The rate of doing work when rope moves with speed of$2.0\mathrm{m}/\mathrm{s}\mathrm{is}225\mathrm{W}$

Step-by-step solution

Given

$$\begin{gathered} \mathrm{W}=900\mathrm{J} \\ \mathrm{speed}=1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Concept

The rate at which force does the work on the object is called as power due to the force. The work done on a particle by a constant force during its displacement is given as

$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

Formula:

$\begin{array}{rcl}\mathbf{W}& \mathbf{=}& \overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}\\ {\mathbf{P}}_{\mathbf{avg}}& \mathbf{=}& \frac{\mathbf{W}}{\mathbf{∆}\mathbf{t}}\\ & \mathbf{=}& \mathbf{Fv}\end{array}$

(a) Calculate the work done

The skier and the rope are moving with constant speed. Hence, the net force on the skier is zero. The force oftherope is balanced by the component of the gravity along the incline. This force is constant and not related to the speed of the rope. Hence, when the same force acts on the rope and moves it throughthesame distance, the work done by the force of the rope isthesame even though the speed changes to $2.0\mathrm{m}/\mathrm{s}$.

The work done is given by

$\mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}$

(b) Calculate the force

$\mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}=\mathrm{Fd}$

Hence, we determine the force as

$\mathrm{F}=\frac{\mathrm{W}}{\mathrm{d}}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{F}& =& \frac{900\mathrm{J}}{8\mathrm{m}}\\ & =& 112.5\mathrm{N}\end{array}$

Hence the force is, 112.5 N.

(c) Calculate the average power

Here, both the force and velocity of the rope are in the same direction. Hence, the rate of doing work i.e. power can be calculated as

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\frac{\mathrm{W}}{∆\mathrm{t}} \\ {\mathrm{P}}_{\mathrm{avg}}=\mathrm{Fv} \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& 112.5\mathrm{N}\times 1.0\mathrm{m}/\mathrm{s}\\ & =& 112.5\mathrm{W}\end{array}$

When the speed of the rope changes, only the power will change since the force of the rope is constant and independent of speed. Hence we get,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\frac{\mathrm{W}}{∆\mathrm{t}} \\ {\mathrm{P}}_{\mathrm{avg}}=\mathrm{Fv} \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& 112.5\mathrm{N}\times 2.0\mathrm{m}/\mathrm{s}\\ & =& 225\mathrm{W}\end{array}$

Hence the rate of doing work when rope moves with speed of $1.0\mathrm{m}/\mathrm{s}\mathrm{is}112.5\mathrm{W}$and The rate of doing work when rope moves with speed of $2.0\mathrm{m}/\mathrm{s}\mathrm{is}225\mathrm{W}$.