- #1
micaele
- 13
- 0
Homework Statement
A 30.0 kg box sliding at 5.00 m/s on a rough surface is brought to rest by 20.0 N of friction. What distance does the box slide?
Homework Equations
KE = 1/2*m*(v^2-v^2)
The Attempt at a Solution
I know that the change in velocity is 5.00 m/s since it is going to rest, I understand the weight and how it works into the equation... but where the Newtons come into play and how I use this to find the change in distance is where I am lost.
I know that Joules are Newton-meters (N*m), but I'm not too sure on converting.
If I put 30.0 kg and 5.00 m/s into the above equation I get KE = 0.5*30.0 kg*5.00 m/s = 375 J = 375 N*m