Under the action of a force, a 2 kg body moves such that its position

1.	Under the action of a force, a 2 kg body moves such that its position x as function of time is given by x = 2t^(4)+ 5, where x is in metres and t is in seconds. The incase in kinetic energy one second after the start of nation is (A) 8J (B) 16J (C) 32J (D) 64J
Answer» ${\mathbb{\underline{\huge{\pink{Hello...}}}}}$ $\bf\huge\boxed{Answer}$ 16 J Explaination :- Position of the body VARIES as X= $\frac{ {t}^{3} }{3}$ Thus the acceleration of the body at any INSTANT= $\frac{ {d}^{2}x }{d {t}^{2} } = 2t$ FORCE any on body at any instant = ma = 2mt Thus work done on body in moving the object by a distance dx = F.dx = 2mtdx = $2mt( {t}^{2} dt)$ Thus TOTAL work done in 2s= $2s {∫}^{2} _{0}2m {t}^{2} dt = 16 \: J$ $\large{ \underline{ \bold{Thank \: my\: Answer}}}$