# Box sliding down an inclined plane

## Homework Statement

A 2kg box is at the top of an inclined plane that is 10m long. The inclined plane makes an angle of 30° with the ground. The μk of the ramp is 0.36 and the μk of the flat ground at the end of the ramp is 0.50. How far does the box travel along the flat ground until it stops?

I'm not quite sure what to do after finding the acceleration, but I think you need to find the final velocity of the box when it finishes going down the ramp? I can't seem to figure out what to do past this point, so any help would be greatly appreciated. Thanks for your time.

## The Attempt at a Solution

1. Fg=mg
Fg=2kg(10m/s^2)
Fg=20N

2.Fgx=mgsinθ
Fgx=2kg(10m/s^2)sin30°
Fgx=10 N

3. Fgy=mgcosθ
Fgy=2kg(10m/s^2)cos30°
Fgy=17.32N

4.a=g(sinθ-μkcosθ)
a=10m/s^2(sin30°-0.36cos30°)
a=1.88m/s^2

Vf^2=(0m/s)^2 +2(1.88m/s^2)(10m)
Vf=6.13m/s

## The Attempt at a Solution

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Doc Al
Mentor
I'm not quite sure what to do after finding the acceleration, but I think you need to find the final velocity of the box when it finishes going down the ramp?
Sounds good.
I can't seem to figure out what to do past this point, so any help would be greatly appreciated.
Figure out the new acceleration of the box, once it starts sliding on the flat ground.

Figure out the new acceleration of the box, once it starts sliding on the flat ground.[/QUOTE]

How would you find this? Would you use a=F(net)m? If yes, how would you find the applied force?

Doc Al
Mentor
Would you use a=F(net)m?
I assume you mean a = F(net)/m. But yes.
If yes, how would you find the applied force?
What force acts on the box as it slides across the flat ground?

Friction?
So on the flat surface you find the force of friction?
FFk=μk*FN
FFk=0.50*20N
FFk=10N
How would you find the applied force using this though?
With this doesn't the formula a=F(net)/m still leave two variables?

Doc Al
Mentor
Friction?
So on the flat surface you find the force of friction?
FFk=μk*FN
FFk=0.50*20N
FFk=10N
Right.
How would you find the applied force using this though?
There's no 'applied force'. Friction is the only (horizontal) force acting.
With this doesn't the formula a=F(net)/m still leave two variables?
No. You know the mass and the net force.