(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horizontal, by a constant force F with a magnitude of 120N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.25.

a.) What is the work done on the oven by the force F?

b.) What is the work on the oven by the friction force?

c.) Compute the increase in potential energy for the oven.

d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.

e.)Use summation{F} = ma to calculate the acceleration of the oven. Assuming the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 14.0m. Compute from this the increase in the oven's kinetic energ, and compare this answer to the one you got in part d.

2. Relevant equations

W = F * x

U = mgy

K = 1/2(mv^2)

U1 + K1 + Wother = U2 + K2

3. The attempt at a solution

I think there are some typo errors from the back of the book:

a.) What is the work done on the oven by the force F?

W = F * x

W = 120N * (14.0cos(37)) <<<< (x component)

W = 1341.71 J

(the answer at back of book is 1690J)

b.) What is the work on the oven by the friction force?

Ff = ukn = 0.25(12.0kg)(9.8m/s^2) = 29.4N

Wf = Ff ( x)

Wf = (29.4N * 14.0cos(37))

Wf = 328.72J = 329J

(the answer at back of book is 329J)

c.) Compute the increase in potential energy for the oven.

U2 = mgy2 = (12.0kg)(9.8m/s^2)(14.0sin(37)) = 990.8J = 991J

(the answer at back of book is 991J)

d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.

U1 + K1 + Wother = U2 + K2

0 + 0 + (WF - Wf ) = U2 + K2

1341.71 J - 329J - 991J = K2

K2 = 21.71J

(the answer at back of book is 360J)

e.) Use summation{F} = ma to calculate the acceleration of the oven. Assuming the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 14.0m. Compute from this the increase in the oven's kinetic energ, and compare this answer to the one you got in part d.

F - Ff = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)

Vf = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J <<< not same as d i dunno why

(the answer at back of book is a = 0.699m/s^2: K = 360J)

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# Homework Help: Potential/Kinetic Energies

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