00:01
So part d of this problem asks us to find the coefficient of friction along the length of this surface after this one kilogram crate is launched at four meters per second and comes to rest after going 0 .89 meters.
00:16
So this problem is a fairly straightforward conservation of energy problem.
00:22
We're going to start down here with kinetic energy.
00:26
And as this block slides up, because of friction, we are going to start.
00:31
To lose energy because friction is going to do work against this crate.
00:36
Once this block gets all the way to the top, it's going to stop.
00:40
So all of this energy, all of the energy that's left over is going to turn into gravitational potential energy.
00:49
So from here, we should see that the work done is because of friction, and friction has this coefficient built in.
00:56
So if we expand on the work done by friction, then we'll be able to solve for the coefficient of friction.
01:01
So the initial kinetic energy is just one -half mv squared.
01:07
The work done by friction is the friction force times the displacement, and we know that this block is sliding a distance of 0 .89 meters just before it hits the spring.
01:19
And once it gets to the top, all of this energy is going to convert into gravitational potential energy, which is mgh.
01:27
So we can expand on this a little bit further.
01:31
We're not going to do anything with the kinetic energy just yet, but the friction force we should know is mu times the normal force, which is good because now we just expanded this equation to include the variable that we're looking for.
01:48
And here is still mgh, we didn't do anything with that.
01:51
And for an object on an incline, we should know that the normal force is going to be mgh, m .g.
02:04
Cosine theta.
02:05
This is the normal force.
02:07
I'm going to assume that you know how to find this already.
02:11
So this is the work done by friction.
02:14
Mew is the coefficient of friction.
02:17
Mg.
02:17
Cosine theta is the normal force...