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How do I solve this question about pushing a box up a ramp?

If you could, please put the steps in as well.

Thanks in advance!

How do I solve this question about pushing a box up a ramp? If you could, please put-example-1

1 Answer

5 votes

Answer:

See below

Step-by-step explanation:

Translating the weight of the box , mg, to the plane coordinates

Fdp = force downplane = mg sin 20.8

Fn = mg cos 20.8 so Ff = Force of friction = .0704 mg cos 20.8

SO the total force acting to slow the box is the sum of these two

= mg (.0704 cos20.8 + sin 20.8°) = 37(9.81) ( .4209) = 152.78 N

The initial KE of the box = 1/2 (37)(7.15)^2 = 945.77 J

Remember Force X distance = Work

the forces slowing the box must do this much work to stop it

152.78 * d = 945.77

d = 6.19 m

User David Hay
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