00:01
In this following question, it is said that two workers are sliding 410 kz crate across the floor.
00:08
One worker pushes forward on the crate with a force of 430 newton while other pulls in the same direction with a force of 270 newton.
00:17
Using a rope connected to the crate, both forces are horizontal and the crate slides with a constant speed.
00:24
Then what is the crate's coefficient of kinetic friction on the floor? so, this is the free body diagram of given activity shown here.
00:35
And since crid is moving with constant velocity, so force in x direction is 0 and force in y direction is also 0.
00:57
Then r will be equal to nw which is equal to mg.
01:02
So from here r is equal to 410 into 9 .8 which is equal to 4018 newton and assume this equation as first equation.
01:16
And also 430 plus 270 is equal to the frictional force.
01:25
Then 430 plus 270 will be equal to mu k into r.
01:32
So putting the value of r, this will be 700 equal mu k into 4018.
01:42
So from here, the kinetic frictional force is equal to 0 .174.
01:52
And in the part b, it is said that a 1600 kg car traveling at its speed of 11 meters per second skits to a halt on weight concrete where kinetic friction is given.
02:03
How long are there skid marks? so this is the figure according to the question.
02:11
And here we will assume s equal length of skid marks...