00:01
Okay, in this question there is an incline surface, okay, at the angle that is theta equals to 37 degree here and a mass of 2kg, okay, that is at rest on the inclined pane, okay, is pulled up by the applying a constant force.
00:16
F, that is, f is given 20 newton parallel to incline, okay? and force act for one second, so t is given one second.
00:25
And now part of the question is show that the work done by the applied force, force, does not exit 40 june okay so we can say here it is f okay so m g will be for the downward so the the component the perpendicular component it will be m g sign theta okay and it is going upward so we can say it will going upward here so it will going m a okay so result end will be m a that is equal to f minus m g sine theta okay so from here f will be m a plus m g sine theta.
01:09
Now we can find out a here.
01:14
We have to find out acceleration.
01:15
So ma will be f minus mg sine theta.
01:18
Now we will put values.
01:21
So ma that is equals to f that is given 20 newton minus mg.
01:30
M is 2 and as per the question g is 10 meter per second squared...