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# im regretting picking honor physics..please help...?

A box with mass 6.6 kg (starting at point A) slides down a ramp inclined at 35 degrees. It was released from rest 3.25 meters above the ground. Near the bottom of the ramp, the box hits a spring of length 1.1 meters with a k of 5.8 N/m which is compressed 0.26 meters (point B). What is the height of the box at B? How much work does friction do from point A to point B? What is the distance from point A to point B along the ramp? What is the coefficient of friction between the ramp and the box?

At this point, the spring releases and the box moves back up the ramp (to point C). How high above the ground is the box when it stops?

### 2 Answers

- NCSLv 73 years agoFavorite Answer
"What is the height of the box at B?"

Initially, height hA = 3.25m, given.

Assuming that the spring lies along the bottom 1.1 m of the ramp, then at maximum compression (point B) the height is

hB = (1.1m - 0.26m) * sin35º = 0.48 m

so the loss in gravitational potential energy is

ΔGPE = mgΔh = 6.6kg * 9.8m/s² * (0.48 - 3.25)m = -179 J

The increase in spring potential energy is

ΔSPE = ½kx² = ½ * 5.8N/m * (0.26m)² = 0.196 J

EDIT after your comments (very helpful):

These numbers don't make much sense. They suggest that friction had drained almost all of the energy from the system by the time the spring is fully compressed.

friction work = 179 J = µ*m*g*cosΘ*d

179 J = µ * 6.6kg * 9.8m/s² * cos35º * (3.25m - 0.48m) / sin35º = µ * 256J

µ = 0.70 ◄

At maximum compression, the spring force is

F = kx = 5.8N/m * 0.26m = 1.51 N

The downslope component of the weight is

W' = mgsinΘ = 6.6kg * 9.8m/s² * sin35º = 37.1 N

so the spring can't push the block back up the incline at all.

It looks to me like it is stuck at point B.

Hope this helps!

- Jim MoorLv 73 years ago
Honor physics is the wrong class, take standard physics.

But you will still need a good working understanding of math

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Does this new solution work?