00:01
In this problem a small block of mass m slides to the left on a frictionless surface with speed v.
00:11
Then the block slides up a rough ramp that has inclined angle of theta.
00:20
The coefficient of kinetic friction between the ramp and the box is mu.
00:25
And we have to find out the height of the block when its speed up the ramp equals kv.
00:34
So the force acting on the block when it is sliding up the ramp will be normal force perpendicular to ramp, gravitational force in the vertically downward direction and friction force will be down the ramp because the block is sliding up the ramp.
01:01
So friction force is opposite to the velocity of the block.
01:06
Now the angle is theta and the direction of the gravitational force with the direction perpendicular to incline will also be theta.
01:19
So the component of gravitational force perpendicular to incline is mg cos theta and so the component of gravitational force parallel to incline is mg sin theta.
01:35
So we have two components of gravitational force.
01:39
Now along the direction perpendicular to incline there is no acceleration so the forces along this direction will balance which means the normal force will be equal to mg cos theta.
01:56
So friction force is coefficient of friction times normal force which is mu mg cos theta.
02:05
Now we consider the motion along the incline.
02:08
So along the incline we have two forces f and mg sin theta.
02:14
So the net force is friction force plus mg sin theta and net force is equal to acceleration sorry net force is equal to mass times acceleration.
02:27
Friction force is mu mg cos theta.
02:38
So mass cancels out and we get acceleration along the incline equal to g sin theta plus sin theta plus mu cos theta.
02:59
This is the acceleration and it will be down the incline opposite to the velocity...