Question
Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20∘20^\circ with the horizontal. (See Figure 7.34.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
<b>Figure 7.34</b> A man pushes a crate up a ramp.
Figure 7.34 A man pushes a crate up a ramp.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

3.14×103 J3.14 \times 10^3 \textrm{ J}

Solution video

OpenStax College Physics, Chapter 7, Problem 5 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to calculate the total work done by this person in moving themselves up the ramp as well as pushing the crate up the ramp. So the work that they do to push the crate up the ramp equals the force they exert on the crate which is 500 newtons along the ramp and then the displacement along the ramp. So that's this term here. Then the work that the person does on themselves is going to be their change in gravitational potential energy which will be their weight multiplied by their change in height and that is the vertical height that we're concerned with there. So the vertical height we can find by multiplying the length of the ramp by sine theta because we're looking for this opposite leg of this ramp triangle. So we substitute d sine theta in for h here, and we get the work done by the person is the mass of the person times g, times d sine theta plus the force exerted on the crate multiplied by the length of the ramp. So we have 85 kilograms times 9.8 newtons per kilogram times four meters times sine 20 degrees, plus 500 newtons exerted on the crate multiplied by the four meter length of the ramp giving us 3.14 times ten to the three joules of work done by the person.

Comments

Why do we only calculate the distance the person moves in height and not along the ramp?

Hi there, thanks for the question. Since there's no friction we assume we're dealing only with conservative forces - which is to say forces (just gravity in this case) where the work done by it depends only on the start and end positions, not on the path taken to get between the start and end points. That means, when working with gravity (or with the electrostatic force ) we can use either the distance along the ramp or the height - those are just different paths to get between the start and end points. We can choose which ever is more convenient. For the sake of understanding you could try using the height change for the crate and multiply that by the crates weight. You can calculate the crate weight by noticing that the force due to the person is equal to the magnitude of the component of gravity along the ramp, in which case the weight of the crate is Fpersonsin⁡(θ)\dfrac{F_\textrm{person}}{{\sin(\theta)}}, which, after multiplying by the change in height of the crate dsin⁡(θ)d \sin(\theta) works out to FpersondF_\textrm{person}d as shown in the video.
Hope this helps,
Shaun