00:01
Suppose the ski patrol lowers a rescue sled carrying an injured skier with a combined mass of 80 .1 kilograms down a 60 degree slope at constant speed, as shown in the given figure.
00:14
The coefficient of kinetic friction between the sled and the snow is 0 .1, so very small, as you would expect, as snow is quite slippery.
00:22
How much work in jules is done by friction as the sled moves 31 .9 meters along the hill.
00:34
So the work is given by the dot product between the force that is applied and the distance traveled.
00:53
So first, let's find the friction force.
00:57
So the friction force is always given by the product of the kinetic coefficient kinetic friction friction times normal force.
01:05
The normal force is not drawing on our illustration, so let's add that.
01:15
To solve for normal force, we're going to use the sums of the forces along this direction equal to zero.
01:23
And we see that n will be equal to the weight of our interest gear times the coast of 60.
01:41
I'll write in theta for now, but this angle right here is 60 degrees.
02:00
So we have the force of friction equal to the coefficient of kinetic friction times the weight of our skier, which we have the mass.
02:14
So we can write m g times the cost of theta where theta is 60 so now we have an expression for our friction force the work will be given by fs times the distance traveled which in this case is 31 .9 meters so now let's input our our numbers for the friction force so 0 .1 for the kinetic friction 80 .1 kilograms for the mass of our skier, including the carrion sled, times 9 .8 for gravity times the coast of 60 degrees.
03:37
And we obtain 1 ,252 joules.
03:48
Now we want to know how much work in jewels is done by the rope on the sled over this distance.
03:56
So now we want to find an expression for, the rope.
04:11
So because we're traveling on a constant speed, what this means that the sum of the forces along the y direction is equal to zero, where i think it's more practical, or excuse me, x, i think it's more practical to incline our system.
04:28
So before it is sums the force on y equal to zero to obtain this expression...