00:01
So we have this toy car and it's going over a rug.
00:03
And we are first told that the mass of the toy car is 0 .24 kilograms.
00:11
And the car is initially moving at a speed of 3 .7 meters per second.
00:21
And then it rolls over a rug.
00:22
And then after it leaves the rug, it is then moving at a speed of 2 meters per second.
00:29
You want to know how much work did the rug do on the car? so let's just draw this scenario out.
00:35
We have smooth, and then we have the rug, and then we're back to smooth.
00:43
And right here, we're moving at that speed v -0, and then when we leave the rug, the little toy truck is moving at some speed v.
00:56
F.
00:57
And so we want to know how much work did the rug do on the car? we know that work is equal to the change in kinetic energy, and this would imply that this is the final kinetic energy minus the initial.
01:15
So that would be one -half m vf squared minus one -half m v -not squared.
01:26
So let's just plug those numbers in.
01:28
So we have work is equal to one -half, and i factor out the m.
01:36
Times the final squared, so 2 meters per second squared, minus v .0 squared, so 3 .7 meters per second squared.
01:47
And we find that the work done by the rug on the car is equal to negative 1 .16 .28 joules.
02:04
And so next we want to know if we're given that the coefficient of friction is 0 .2, how far does the car move across the rug? well, we have to realize that work is equal to force times distance times cosine of theta, where theta is the angle between the force factor and the displacement vector.
02:36
And so let's look at what's going on.
02:38
In this little green portion where we have the rug, we have our toy car still right, and it's moving towards the right.
02:47
That's our displacement vector.
02:49
But the only force acting in that same direction is force of friction, which is opposing that movement.
02:57
Of course, we have normal force going up and weight force going down, but those forces are perpendicular to our displacement vector, and so we don't care about the work from those.
03:08
Because cosine of 90 degrees is zero.
03:12
But cosine the theta between the displacement vector and the force of friction vector is 180.
03:20
So the work, due to force of friction, the only one that we care about, will be equal to frictional force times whatever distance d it went times cosine of 180 degrees.
03:39
And so this is just negative 1, right, which will help us cancel out the negative we got in the first part...