A 260 kg crate hangs from the end of a L = 14.0 m rope. You push horizontally on

Question

A 260 kg crate hangs from the end of a L = 14.0 m rope. You push horizontally on the crate with a varying force F to move it 4 m to the side (Fig. 7-37).

(a) What is the magnitude of F when the crate is in this final position?
(b) During the crate's displacement, what is the total work done on it?
(c) During the crate's displacement, what is the work done by the weight of the crate?
(d) During the crate's displacement, what is the work done by the pull on the crate from the rope?

(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Explanation / Answer

from the given figure

m = 260 kg

L = 14 m

d = 4 m

F = varying

a. when the crarte is at final position

from froce balance

F = Tcos(theta)

Tsin(thewta) = mg

F/mg = 1/tan(theta)

F = mg/tan(theta) = mg*d/sqrt(L^2 - d^2)

F = 760.44199778 N

b. total work done on crate = chang ein PE = mgh

h = L - sqroot(L^2 - d^2)

hence

W = 1488.51 J

c. work done by weight of crate = -W = -1488.51 J

d. work done by the pull form the rope = Wp

Wp = 0 J as force of tension is always prependicular to the motion of the crate

e. work done by F on crate = 1488.51 J ( from work energy theorem)

f. total work is not answer from a * horizontal displacement as force is varying with time

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.