Question

A crate of mass 11.0 kg is pulled up a rough incline with an initial speed of 1.40 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 19.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.94 m. (a) How much work is done by the gravitational force on the crate? J (b) Determine the increase in internal energy of the crate–incline system owing to friction. J (c) How much work is done by the 102-N force on the crate? J (d) What is the change in the kinetic energy of the crate? J (e) What is the speed of the crate after being pulled 4.94 m? m/s

Answer 1

(a) How much work is done by the gravitational force on the crate

W_g = mgdcos (90 + 19)

W_g = 11 * 9.8 * 4.94 * cos 109

W_g = - 173.37 J

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(b) Determine the increase in internal energy of the crate–incline system owing to friction

this is same as work done by friction ( note that friction acts opposite to motion)

W = E = umg* cos 19 *d

E = - 201.4 J

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(c) How much work is done by the 102-N force on the crate

W = 102 * 4.94 * cos 0

W = 503.88 J

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What is the change in the kinetic energy of the crate

using work energy energy

net work done = change in kinetic energy

change in kinetic energy = - 173.37 - 201.4 + 503.88

change in kinetic energy = 129.1 J

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What is the speed of the crate after being pulled 4.94 m

again use work energy theorem

1/2 * 11 * v_f² - 1/2 * 11 * 1.40² = 129.1

v_f = 4.64 m/s