What is the student's speed at the bottom of the inclined plane?

[ConfusedStudent]

Hi, everyone...we just got a test back and I'm correcting my problems so I have the right answers for the final...

An inert physics student slides down a plane inclined at 20 degrees above the horizontal. He feels a constant retarding friction force of 100 N as he slides 10 m down the plane. What is the student's speed at the bottom of the plane?

I'm not going to bother to show you guys my work since I got the question wrong :(

Thanks all

 

[NateTG]

There's not enough information to answer this.

 

[PrudensOptimus]

Assuming the skater started out with 0 initial velocity then the final velocity is:

v = 2sqrt(100μ_k)

 

[NateTG]

Originally posted by PrudensOptimus
Assuming the skater started out with 0 initial velocity then the final velocity is:

v = 2sqrt(100μ_k)

*blink*
We know that the retarding force is 100 N but have no idea what the student's mass is, so we have no idea about inertia or gravity. I'd like to see the work for this one.

 

[NateTG]

Missing information:
Student's mass

Also, is that 10m vertical drop, or 10m along the slide?

 

[chroot]

Nate,

I concur. You also need to know either the mass, or the coefficient of kinetic friction.

The easiest way to solve the problem is via th conservation of energy:

kinetic energy at the bottom = potential energy lost in the fall - work done by friction during the fall

1/2 m v² = d m g sin 20 - d μ_k m g

- Warren

 

[ConfusedStudent]

I'm so sorry, I somehow left out the mass, it was given to be 120kg...

So sorry

Also that is 10m down the slide,.

 

[chroot]

You should have everything you need in my last post -- can you solve for v?

- Warren