Question

the displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time T is given by X equal tq by 3 where X is in metre and T is in second find the work done by the external agent for the first one second​

Answer 1

x = 2 t^2

dx / dt = v = 4t

d2x / dt2 = a = 4

Force acting on a body = m*a = 1 * 4 = 4 Newton

W = \int F . dx

w = \int 4 * 4t dt

= 16 t^2 / 2 = 8 t^2

now put t = 1

hence w = 8 joules

Answer 2

Answer:

x = 2 t^2

dx / dt = v = 4t

d2x / dt2 = a = 4

Force acting on a body = m*a = 1 * 4 = 4 Newton

W = \int F . dx

w = \int 4 * 4t dt

= 16 t^2 / 2 = 8 t^2

now put t = 1

hence w = 8 joules