00:01
Hi, here in this given problem, mass of the crate which is being pulled by the worker in the warehouse, that is m is equal to 90 .0 kilogram force applied by him on the crate, that is 288 newton speed with which the crate is moving 0 .850 meter per second.
00:41
Now, the length of rough surface over which it has to be moved, let it be represented by x and x is 0 .75 meter.
01:02
Coefficient of kinetic friction over that surface, 0 .357.
01:08
So in the first part of the problem we have to find net force acting over the crate, over the net surface, over the, uh, the rarrow.
01:17
Surface.
01:18
So that net force will be given by force applied minus force of friction, kinetic friction, fk.
01:34
So we put the values for the force applied, that is simply f288 for the force of friction that is mu k times the normal reaction which is m g.
01:53
Means this is 288 minus 0 .357 into mass of the crate 90 kilogram into g 9 .8.
02:05
So this net force will be given by 288 minus 314 .9.
02:16
Newton means force of friction exceeding the force applied.
02:20
This is not possible.
02:21
Here this net force is coming to be equal to minus 26 .9 newton...