A force F = 20 + 10 y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle form y = 0 to y = 1 m is

A force F = 20 + 10 y acts on a particle in y-direction where F is in

1.	A force F = 20 + 10 y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle form y = 0 to y = 1 m is
Answer» `20 J` `30 J ` `5 J` `25 J` Solution :GIVEN : `F = 20+ 10 y` work done, `W = int F.DY` `= int_0^1 (20 + 10 y) dy = [20 y + 10/(2 y^2)]_(0)^(1) = 20 + 10/2 = 25J`.