Question Video: Solving a Real-World Problem Involving Work Done by a Constant Force | Nagwa Question Video: Solving a Real-World Problem Involving Work Done by a Constant Force | Nagwa

Question Video: Solving a Real-World Problem Involving Work Done by a Constant Force Mathematics

A man of mass 94 kg ascended a plane of length 90 m which was inclined at an angle of 30ยฐ to the horizontal. Determine the work done by his weight to the nearest joule. Take ๐‘” = 9.8 m/sยฒ.

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Video Transcript

A man of mass 94 kilograms ascended a plane of length 90 meters, which was inclined at an angle of 30 degrees to the horizontal. Determine the work done by his weight to the nearest joule. Take ๐‘” to equal 9.8 meters per second squared.

Okay, so letโ€™s say that this is our plane at 30 degrees to the horizontal. And thereโ€™s a man of mass 94 kilograms climbing up the plane. Weโ€™ll call that mass ๐‘š. And weโ€™re told that the length of the plane is 90 meters, which weโ€™ll call ๐‘™. We want to calculate the work done by this manโ€™s weight to the nearest joule.

We can say then that this manโ€™s weight, his mass times the acceleration due to gravity, a downward-acting force, actually does work as this man climbs up the incline. That work is equal to the force involved, that is, the manโ€™s weight force, multiplied by the vertical distance through which the man moves. That is, the distance weโ€™ll use in our calculation wonโ€™t be the length ๐‘™. But rather it will be this vertical distance ๐‘‘. Thatโ€™s because this distance represents the only component of the length ๐‘™, we could call it, that is parallel or antiparallel to the force involved.

Hereโ€™s where we can start then. We know that the force for which we want to calculate work done is the manโ€™s weight force, ๐‘š times ๐‘”. Going a step further, since this 30-degree slope is part of a right triangle, we can say that the distance ๐‘‘ is equal to ๐‘™ times the sin of 30 degrees. This is so because the sine of this angle is equal to the ratio of ๐‘‘ to ๐‘™. Rearranging that relationship, we get that ๐‘‘ equals ๐‘™ sin 30.

Before we plug in the given values for ๐‘š, ๐‘”, and ๐‘™, itโ€™s important to set up a sign convention. Note that as this man climbs the incline, he moves upward vertically. This is in the opposite direction to his weight force, which acts down. To make sure that we combine quantities correctly, letโ€™s say that one of these directions is positive, making the other negative. If we say that the upward direction is positive, then that means the value of ๐‘™ times the sin of 30 degrees is positive. But then that means that ๐‘š times ๐‘”, which is directed downward, must be negative. The point here is that the force involved, the manโ€™s weight force, is acting in an opposite direction to the displacement ๐‘‘. The net effect of this is that the work done by this weight force will be negative.

Now that weโ€™ve figured out the signs, weโ€™re ready to substitute in for ๐‘š, ๐‘”, and ๐‘™. The manโ€™s mass is 94 kilograms, ๐‘” is 9.8 meters per second squared, and ๐‘™ is 90 meters. Entering this expression on our calculator, we get a result, to the nearest joule, of negative 41,454 joules. This, to the nearest joule, is the work done by this manโ€™s weight as he ascends this plane.

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