Chaya Galloway

2021-02-12

A skier of mass 70 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him 60m up a 30 degree slope (assumed frictionless) at a constant speed of 2.0 m/s ?(b) What power must a motor have to perform this task?

pattererX

Skilled2021-02-13Added 95 answers

Mass of skier (m)=70 kg

Applied Force (F)=$mg\mathrm{sin}30=345.35\text{}N$

He must be pulled up d=60m to a height of 30 meters by doing the work (W).

slope (assumed frictionless)

$W=mg\mathrm{sin}30\cdot d$

$W=20.6\text{}kJ$

(b) The constant speed (v)=2.0 m/s

The power of motor $\left(P\right)=F\cdot v$

P=686.70W

Jeffrey Jordon

Expert2021-10-14Added 2575 answers

We know that, $W=Fd\mathrm{cos}\theta $

We must identify the force that is pulling the skier up the hill a distance of d in order to calculate the work that was required to do so.

The force in this situation is the cable T Tension.

Note that, the skier is moving at a constant speed. This means that the forces acting on him in the x-direction is equivalent.

From the figure above you may notice that the two forces that are acting on him in the x-direction are, the x-component of his own weight and the tension in the cable.

We analyzed the weight of the skier into its x-y components, as you see in the figure above, to make it easy to find the value of the tension.

We are going to use Newton's second law to find the value of the tension T.

$\sum {F}_{x}=T-mg\mathrm{sin}\theta =m{a}_{x}$

Note that ${a}_{x}=0$ because the speed is constant

$T-mg\mathrm{sin}\theta =m{a}_{x}=0$

$T-mg\mathrm{sin}\theta =0$

$T=mg\mathrm{sin}{30}^{\circ}$

From the work law, we know that, $W=Fd\mathrm{cos}{\theta}_{2}$

Note that the angle ${\theta}_{2}$ is the angle between the tension and the displacement d

$W=Td\mathrm{cos}{0}^{\circ}$

Note that, ${\theta}_{2}={0}^{\circ}$ because the tension and the displacement are in the same direction

$W=Td\text{}\text{}\text{}\text{}\text{}\text{}(\mathrm{cos}{0}^{\circ}=1)$

$W=Td=[mg\mathrm{sin}{30}^{\circ}]d\text{}\text{}\text{}\text{}\text{}(Substituting\text{}T\text{}from(1))$

$W=mg\mathrm{sin}{30}^{\circ}d=mgd\mathrm{sin}{30}^{\circ}$

$W=mgd\mathrm{sin}{30}^{\circ}=70\times 9.8\times 60\mathrm{sin}{30}^{\circ}$

$W=2.1\times {10}^{4}\text{}J$