A 25 kg air compressor is dragged up a rough incline from r₁ (→ a...

Question

A 25 kg air compressor is dragged up a rough incline from r₁ (→ above r)= (1.3î + 1.3ĵ) m to r₂ (→ above r) = (8.3î + 2.9ĵ) m, to where the y-axis is vertical. How much work does gravity do on the compressor during this displacement?

Accepted Answer

Hi everyone in this particular practice problem. We are asked to actually determine the work done by the gravitational force during the rescue mission where we'll have a search and rescue helicopter lifting, a distressed sailor with a mass of 85 kg using a cable from point A with a car coordinate of X A equals 3.5 times to the power of three m and A Y A of zero m to point B with a car coordinate of X B equals 2.2 times 10 to the power of three m and Y B equals 18.5 m. And we're asked to calculate the work done by the gravitational force during this rescue mission. Assuming that the sailor is lifted up at a constant speed. The options are minus 1.1 times 10 to the power of six Jule minus 1.54 times 10 to the power of four Jule, 1.54 times 10 to the power of four Jule and 1. times 10 to the power of six Jule. So because the sailor has been lifted with a constant speed, the gravitational force F G or M G acting on the sailor has a constant magnitude and a constant direction. So F G can equals to minus M G in the J direction or in the direction pointing towards the center of the earth or pointing downward. And then when we're being asked to calculate the work done by the gravitational force during this rescue mission, that will equals to um finding the dog product of its weight or its gravitational force and the displacement. So this will equals to F G dot product to the delta R. So what we have to find now or what we have to start with is to by cal is to calculate the sailor's displacement by first listing sealer's initial position, which will be R A which will equals to an X component of 3.5 times 10 to the power of three m in the I direction or in the X direction plus zero m in the J direction or in the Y direction. And then the final position is going to be given by R B which will equals to 2.2 times 10 to the power of three m in the X or the I component plus 18.5 m in the J or in the Y direction or component. Using this final and initial position, we can find the sailor's displacement which will be delta R equals R B minus R A which will then be just 2.2 times 10 to the power of three minus 3.5 times 10 to the power of three meter in the I component plus 18.5 minus zero m. In the J component, the delta R will then be minus 1.3 times to the power of three m in the I component N plus 18.5 m in the J component just like. So now that we find the sailor's displacement, we can actually start calculating the work done by the gravitational force by finding dot product of the F G with delta R. So this will give a value of equals um The M G or F G is minus M G which will be minus 85 kg multiplied by 9.81 m per second squared in the G direction. And that will be dot two minus uh 1.3 times 10 to the power of three m in the I component plus 18.5 m in the J component just like. So and then the way we want to do this is to by actually uh multiplying each of the terms here. So the F G will be multiplied by first the um X component in the delta in the delta R and then the Y component as well in the delta R. But what we need to recall is the fact that um based on factor properties, we need to note that J dot product with or the J component dot product with an I component will equals to zero and AJ dot product with AJ component will then equals to one essentially because the gravitational force is only acting on one direction, which is only on the J component. We can neglect what we have in the displacement in the I component here. So that is essentially just saying that the W is going to only be equal to 85 kg multiplied by Uh or -85, multiplied by 9.81 m/s squared. Also multiplied by the J component of the displacement which is 18.5 m J dot product to J which is equals to one and the J dot Product with an I is essentially going to be neglected because it's going to equals to zero. So we can just not write that down and neglect it from the start. This will actually equals two. A work value of -1.54 times 10 to the power of four. And that will be the work done by the gravitational force considering that the gravitational force is going to only be pointing on the constant direction of towards the center of the earth or only on the J component exclusively. So our W or over value of minus 1.54 times 10 to the power of four will correspond to option B. So option B is going to be the answer to our practice problem. And that will be all for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics, especially on how to do a dot product and how to find displacement and that'll be all for this one. Thank you.

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A 25 kg air compressor is dragged up a rough incline from r₁ (→ above r)= (1.3î + 1.3ĵ) m to r₂ (→ above r) = (8.3î + 2.9ĵ) m, to where the y-axis is vertical. How much work does gravity do on the compressor during this displacement?

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