A 200 kg crate hangs from the end of a L = 12.0 m rope. You push horizontally on

Question

A 200 kg crate hangs from the end of a L = 12.0 m rope. You push horizontally on the crate with a varying force F to move it 4.00 m to the side (Fig. 7-37). (a) What is the magnitude of F when the crate is in this final position? 1 . N During the crate's displacement, what are (b) the total work done on it, 2 . J (c) the work done by the weight of the crate, and 3 . J (d) the work done by the pull on the crate from the rope? 4 J (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. 5 J (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Explanation / Answer

A 260 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.
(a) What is the magnitude of F when the crate is in this final position?
N

(b) During the crate's displacement, what is the total work done on it?
J

(c) During the crate's displacement, what is the work done by the weight of the crate?
kJ

(d) During the crate's displacement, what is the work done by the pull on the crate from the rope?
J

(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.
kJ

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

Mass of the crate is m = 260 kg
Length of the rope is L = 14.0m
a) The horizontal component of the ropes tension is T sin
sin = 4.00 / 14.0
= 16.6 degrees
vertical component of the tension supports aganist the weights is
T cos = mg
T = 260 kg * 9.8 m/s^2 / cos 16.6
= 2658.8 N
the magnitude of F when the crate is in this final position is
F = 2658.8 N sin 16.6
= 759.5 N
b) Since there is no kinetic energy there is no work done
c ) The work done by the gravity is
W = F * d
= - mg h
h = ( 1- L cos)
= ( 1 - 14.0 m cos 16.6)
= -12.4m
W = - 260 kg * 9.8 m/s^2 * -12.4 m
= -31.6 k J
d ) The tension vector is perpendicular to the direction of the motion so work done is zero
e) the work your force F does on the crate is W_f = - W_g
= + 31.6 k J
f ) the work of your force not equal to the product of the horizontal displacement because it has no constant magnitude

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