A 230 kg crate hangs from the end of a rope of length L=12.0 m. You push horixon

Question

A 230 kg crate hangs from the end of a rope of length L=12.0 m. You push horixontally on the crate with a varying force F to move it a distance d=4.0 m to the side.

a.) What is the magnitude of F when the crate is in its final position?

b.) The total work done on it?

c.) The work done by the gravitational force on the crate?

d.) 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 b, c, d, to find the work your force F does on the crate?

f.) Why is the work of your force from the previous part not equal to the product of the horizontal displacement, d, that the crate undergoes and the anser to a,?

Explanation / Answer

(A) x = Lsin@

4 = 12 sin@

@ = 19.5 deg

in this position,

in vertical, Tcos19.5 = mg

T = (230 x 9.8) / cos19.5 = 2390.7 N

in horizontal, Tsin19.5 = F

F = 2390.7 sin19.5 = 798 N .....Ans

(B) total work done on system = change in KE

{ work - energy theorem}

and change in KE = 0

hence work done = 0

(C) height moved, h = L(1- cos19.5) = 0.7 m

Work done = - m g h = -230 x 9.8 x 0.7 =- 1551.4 J

(D) work done by pull force ( Tension).

tension is always perpendicular to the motion of crate.

and W = F.d

hence work done by pull force will be zero.

(E)   Work done by all forces = 0

0 - 1551.4 + Wf = 0

Wf = 1551.4 J

(F) becasue force F is varying .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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