00:01
Here, a block has a mass of 6 kilograms and it sits on the incline of 35 degrees.
00:10
First, we have to calculate value of force which is required to stop the block if there is no friction.
00:23
Let's illustrate the forces which are acting.
00:27
Gravity is acting downwards.
00:29
There is a normal reaction up the incline and also the force f which we apply here.
00:39
And as we can notice if we introduce y and x -axis, f equals to m g sine alpha.
01:03
Let's calculate it.
01:13
There is 33 .7 newtons, which is roughly 34 neutons.
01:21
Now let's move on to question b.
01:24
In question b we have to take and take out static friction of 0 .20.
01:31
So now this block, in addition to pulling force f, is experiencing static friction force.
02:07
And now, m .g.
02:10
Sine alpha, equals to static friction force plus f.
02:22
And as we know, static friction force is coefficient of static friction force times fn, which is coefficient of static friction force.
02:34
Efficient m g cosine alpha therefore f equals to m g sine alpha minus coefficient of steady fxifficient force mg cosine alpha and thereby the force f which we must apply is m g times sine alpha minus mu cosine alpha so there's 24 .09 new new neutrons which is roughly 24 new and now let's answer question c.
03:36
In question c we have to calculate value of the speed, sorry, value of the force, which is necessary to keep the velocity constant...