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(a) What is the magnitudeof F when the crate is in this final position? (b) Duri

ID: 1765023 • Letter: #

Question

(a) What is the magnitudeof F when the crate is in this final position?


(b) During the crate's displacement, what is the total work done onit?
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(c) During the crate's displacement, what is the work done by theweight of the crate?
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(d) During the crate's displacement, what is the work done by thepull on the crate from the rope?
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(e) Knowing that the crate is motionless before and after itsdisplacement, use the answers to (b), (c), and (d) to find the workyour force F does on the crate.


(f) Why is the work of your force not equal to the product of thehorizontal displacement and the answer to (a)?
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Explanation / Answer

(a) horiz forces:     F =Tsin               vert forces:     mg =Tcos   .          divide:             F / mg = tan           where   sin = 4.00 / 12.00     = 19.47 degrees .      F = mg tan = 230 * 9.80 * tan19.47 =   797 Newtons . (b) total work done = change in kinetic energy = zero . (c) work done by mg is -mgh = - 230 * 9.80 *(12 - 12cos19.47) =    -1547 Joules . (d) work done by rope is zero, since it is alwaysperpendicular to the motion. . (e)    work done by F   +  work done by gravity = 0 .                   work done by F = 1547 J . (f)   Because the force varied, so the total work isnot simply the distance times the final force.