00:01
In this problem a block of ice released from rest at the top of 1 .50 meter long frictionless ramp slides downhill reaching a speed of 2 .50 meter per second at the bottom.
00:18
Using this we have to find out the angle between the ramp and the horizontal.
00:30
So let us suppose the angle is theta.
00:33
Now the force acting on the block will be normal force by the ramp in the direction perpendicular to ramp and gravitational force in the downward direction.
00:52
Now the angle of gravitational force with the direction perpendicular to ramp will be theta which is same as the angle of the ramp with the horizontal.
01:04
So the component of gravitational force perpendicular to ramp will be mg cos theta and the component of gravitational force parallel to ramp will be mg sin theta.
01:19
Now along the ramp we have only one force mg sin theta so acceleration along the ramp will be force upon mass.
01:31
So acceleration is g sin theta.
01:35
Now we will use the equation of kinematics vf square equal to vi square plus 2as for motion along the ramp.
01:48
Final velocity is 2 .50 meter per second.
01:55
Initial velocity is zero at the top of the ramp.
02:00
Acceleration is g sin theta and s is the displacement.
02:06
From this we get sin theta equal to 2gs...