Reviewing for my exam (Newton question)

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The discussion centers on a physics problem involving the motion of a 20 kg box pushed up a hill with a force of 200 N over a distance of 6 m, while overcoming a frictional force of 100 N. The calculations reveal that the box reaches an elevation of 2.86 m when its speed is 2 m/s. Additionally, the angle of the hill with the horizontal is determined to be 25.5 degrees. The energy conservation principle is applied to derive these results, confirming the correctness of the calculations presented.

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Homework Statement



You push a 20 kg box from an elevation of zero up a hill. You exert a force of 200 N for 6 m and the box speeds up steadily from rest to 2 m/s. There is a 100 N frictional force that opposes the box’s motion.

A) Draw energy bar charts to describe this motion. Be sure to describe your system, and the initial and final states.

B) At what elevation is the box moving when its speed is 2 m/s?

C) What angle does the hill make with the horizontal?






A) Initial (Bottom) Work = Final (Top) PE + KE+ [tex]\Delta[/tex]Uinternal

B) Givens

m=20kg
F-applied = 200N
F-friction = 100N
V = 2m/s
L(distance traveled) = 6m

H= ?

Fd = mgh + .5mv2 + [tex]\Delta[/tex]Uinternal

(200N)(6m) = (20kg)(9.80m/s^2)h + .5(20kg)(2m/s)^2 + (100N)(6m)

1200J = 196h + 40J + 600J

600J = 196h + 40J

560J = 196h

H = 2.86m


C) tan[tex]\theta[/tex] = 2.86/6 = 25.5 degrees

Not sure if anything is correct to be honest.
 
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Actually I think that's about right. Just make sure you're using the right trigonometric function for the last step.
 

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