00:02
In this problem, we're given a 3 .5 kilogram block that travels over a smooth surface that i have highlighted in blue, and then it transitions to a rough surface that i've highlighted in orange with a kinetic coefficient of friction of 0 .25.
00:24
It's connected via a spring that has a spring constant of 640 newtons per meter, and the block has a mass of 3 .5 kilograms.
00:36
Now, for the first part, we need to find out the total thermal energy dissipated as the block travels a distance d, and d is given as 7 .8 meters, so d is equal to 7 .8 meters, and then it comes to rest.
00:54
So for that, we need to calculate the frictional force exerted when the block reaches the orange surface.
01:03
The kinetic friction is given by f equals neo sub k times the contact force r.
01:11
Now let's consider the forces acting on the block vertically.
01:17
So it's being pulled down by the weight that is equal to the mass times gravity, and it's being pushed up by the contact force are.
01:31
Now, there is no net vertical acceleration, so the contact force must be equal to the weight.
01:40
That means that r equals mg.
01:44
And we can go ahead and make the substitution in our first formula.
01:48
So that gives f equals muz of k times m g.
01:54
Now note that all of these are constants.
01:58
The coefficient of kinetic friction is a constant.
02:00
The mass of the block is a constant.
02:03
And the gravitational acceleration is also constant.
02:08
Now when we have a constant force, it's work done.
02:13
Is the force times the distance traveled under the influence of the force.
02:21
So the work done is actually equal to mu sub k times m times g times d.
02:30
And we have all of these values so mu equals 0 .25.
02:35
The mass of the block is 3 .5 kilograms.
02:39
Gravitational acceleration is 9 .8 meters per second squared and d is 7 .8 meters now it's a simple matter of performing the multiplication, which we can do using a calculator...