00:01
Okay, in this question a, straight track is there, okay, and the box on this track is there having 2 ,000 newton weight, okay? and it slowly slide through 20 meter, so as displacement is given 20 meter, okay, and the track having frictional coefficient that is mu equals to 0 .2.
00:19
Okay, now in the party of this question, we have to find out the work done by the person pulling the box with a chain at an angle theta with the horizontal.
00:27
So this is the horizontal, okay, and this will be the.
00:30
The theta so the force is applying at this theta angle okay so horizontal component of this force it will be f cost theta and vertical component of this force it will be f sine theta okay and due to the gravitational force there will be 2 ,000 newton downwards okay there will be force and f sine theta will be upward so resultant contact force it will be f that is 2 ,000 newton minus f sine theta okay because 2000 newton is downward and f sine theta is upward.
01:04
So this will be r.
01:05
And f cost theta there will be for right side.
01:10
So equal and opposite, mu r will be for opposite side.
01:15
So f cost theta here it is equal to mu r.
01:20
So f cost theta value of mu is given 0 .2 and r from here 2000 minus f sine theta...