Finding magnitude of force: given v,d,friction coef,mass.

[modded]

Homework Statement

If the 50-kg crate starts from rest and achieves a velocity of v = 4m/s m/s when it travels a distance of 5 mm to the right, determine the magnitude of force acting on the crate. The coefficient of kinetic friction between the crate and the ground is = 0.3

The crate is being pulled by the unknown force(p) at 30 degrees to the right.

Homework Equations

F_n=m*g
F_f=μ*F_n
∑F=m*a
a=dv/dt
v=ds/dt

The Attempt at a Solution

The weight ends up being 491.

I started off with summing up the forces in the x and setting them equal to m*a
∑Fx= cos(30)p-0.3(491)=50*a

Then I found the sum of the force in y and set it equal to 0.

∑Fy= sin(30)p+491=0
p=497

When I plug that back into the x equation I get a=5.7

Then I am not sure what to do next.

Do I use the a=dv/dt next?

Thanks again. I had statics last year and I am currently in dynamics and calc 1 so the integration, derivation, and sum of the forces no longer = 0 have me going crazy. I have never used this forum for help so I hope I set this up clearly for you.

 

[Simon Bridge]

∑Fx= cos(30)p-0.3(491)=50*a

... so "p" is the unknown pulling force and 491N is the weight of the crate... is that correct?
The friction force depends on the coefficient (you got that) and how hard the crate presses into the ground.
The unknown force has an upwards conponent - what does that do to how hard the crate presses into the ground?

∑Fy= sin(30)p+491=0
p=497

That is only the case if the unknown forces upwards conponent is exactly equal to the weight.
Is that the case? Is the crate "floating"? If sin(30)p < 491 - what happens? Are there any other forces acting vertically on the crate?

Note: $\cos(30^\circ)=\frac{\sqrt{3}}{2}$