A 271 kg crate hangs from the end of a rope of length L = 12.6 m. You push horiz

Question

A 271 kg crate hangs from the end of a rope of length L = 12.6 m. You push horizontally on the crate with a varying force to move it distance d = 2.69 m to the side (see the figure). (a) What is the magnitude of when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (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 the answers to (b), (c), and (d) to find the work your force does on the crate.

Explanation / Answer

a)To hold the crate at equilibrium in the final situation,F must have the same magnitude as the horizontal component of the rope’s tension T sin , where is the angle between the rope (in the final position) and vertical:

sin =d/L =2.69/12.6 => =sin-1(d/L)

=12.33o

But the vertical component of the tension supports against the weight:

T cos = mg

T =mg/cos =271*9.8/cos12.33 =2718.5N

F =Tsin =2718.5sin12.33 =580.5N

b) Since there is no change in kinetic energy, the net work on it is zero i.e

Wnet =0

C)The work done by gravity is

WG =-mgh where h =L(1-cos)

WG =-271*9.8*12.6(1-cos12.33) =-771.86J 0r -0.77KJ

d)(d) The tension vector is everywhere perpendicular to the direction of motion, so its work is zero (since cos 90° = 0)

W=0

e)The implication of the previous three parts is that the work due to F is –WG (so the net work turns out to be zero). Thus

WF=-WG =771.86J or o.77kJ

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