A variable force, given by the 2-dimensional vector F = (3x^2 hati + 4hatj), acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0)? (The coordinates are in metres)

A variable force, given by the 2-dimensional vector F = (3x^2 hati + 4

1.	A variable force, given by the 2-dimensional vector F = (3x^2 hati + 4hatj), acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0)? (The coordinates are in metres)
Answer» `-7J` Zero `+7 J` `+ 19 J` Solution :`vecF = 3x^2 hati + 4hatj` `VECR= XHATI + y hati :. dvec(r) = dx hati + dy hatj` WORK done, `W= int vecF . Dvecr` `= int_((2,3))^((3,0))(3x^2 hati + 4 hatj).(dx hati + dyhatj)` `= int_((2,3))^((3,0)) 3x^2 dx + 4dy = int_((2,3))^((3,0)) d(x^3 + 4y)` `= [x^3 + 4y[_((2,3))^((3,0)) = 3^3 + 4 xx 0 - (2^3 + 4 xx 3)` `= 27 xx 0 - (8 + 12) = 27 - 20= _7 J`. Accroding to work energy theroram, Change in the kinetic energy = Work done , `W = + 7 J`.