vagnhestagn

2022-10-02

A cable pulls an 344 kg safe up a slope inclined at ${24}^{\circ}$, for 37 m on an very slippery oily surface. The safe starts and ends at rest. b) What is the work done by the gravitational force? c) What is the work done by the cable?

tiepidolu

Beginner2022-10-03Added 8 answers

Given data:

Mass of safe, m=344 kg

Angle of inclination, $\theta ={24}^{0}$

Distance, d=37 m

To determine:

b) Work done by the gravitational force.

c) Work done by the cable.

Explanation:

b) Work done by the gravitational force can be calculated as follows:

$W=-mgd\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}W=-(344\text{}kg)(9.8\text{}m/{s}^{2})(37\text{}m)\mathrm{sin}{24}^{0}\phantom{\rule{0ex}{0ex}}W=-50734\text{}J$

c) Work done by the cable can be given as:

$W=-(-50734J)\phantom{\rule{0ex}{0ex}}W=50734J$

Answer:

b) Work done by the gravitational force is −50734 J.

c) Work done by the cable is 50734 J.

Mass of safe, m=344 kg

Angle of inclination, $\theta ={24}^{0}$

Distance, d=37 m

To determine:

b) Work done by the gravitational force.

c) Work done by the cable.

Explanation:

b) Work done by the gravitational force can be calculated as follows:

$W=-mgd\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}W=-(344\text{}kg)(9.8\text{}m/{s}^{2})(37\text{}m)\mathrm{sin}{24}^{0}\phantom{\rule{0ex}{0ex}}W=-50734\text{}J$

c) Work done by the cable can be given as:

$W=-(-50734J)\phantom{\rule{0ex}{0ex}}W=50734J$

Answer:

b) Work done by the gravitational force is −50734 J.

c) Work done by the cable is 50734 J.

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