00:01
Hello, this question investigates equilibrium on an inclined plane.
00:09
So we have a plane that is inclined at an angle of 60 to the horizontal, and a victim is being lowered.
00:20
His mass is 90 kilograms, and so his weight is 90 times gravity.
00:26
So this is gravity, not 90 grams.
00:32
We can resolve the weight to get components vertical to the plane, which is 90 g cross 60 and components parallel to the plane 90g sign 60 the motion is downwards the motion is downwards and so friction act in the opposite direction our frictional force is here the coefficient of friction is given a 0 .1 and of course we have not reaction critical r.
01:25
Now we are to find the work done by friction when the body moves a distance of 30 meters.
01:33
Wadden is force of friction times distance.
01:38
The friction of force is given by mu r so wakdan is mu r times distance but considering the vertical equilibrium of this body r equals 90 times gravity times cost 60.
01:55
So our fission is mule times 90 times gravity times cost 60 then all times 30 so this is our r this works out to be 44 .1 newtons times 30 so the entire thing here works down to 1 3m .3 so we first find friction multiply by distance but our friction is new r and that mew is given as 0 .1 the r is given us 90 times gravity times cost 60 and so we arrive at this.
03:19
The work done by the tension since the body is in equilibrium the net force is zero.
03:25
Now this net force we can evaluate it along the direction of motion...