The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as
`x(m)=3t^(2)+4t+5`
Here, t is expressed in second. What is the work done by the force in first 2 seconds?

Explanation

From the displacement equation, one can determine the expression for force acting on the object It has been calculated below
`x=3t^(2)+4t+5`
Differentiating once, w.r.t. time, we get
`(dx)/(dt)=6t+4 or`
`v=6t+4`
Differentiating again, w.r.t. time we get
`(dv)/(dt)=6`
`impliesa=6 m//s^(2)`
So, the force is `F=ma=0.5xx6=3N`
The body m oves in a straight line and positive sign in force and velocity for the interval `0-2` shows that the angle between force and displacement is `0^(@).`
The displacement `S=x_(2)-x_(g)`
`=[3(2)^(2)+4(2)+5]-[3(0)^(2)+4(0)+5]`
`=12+8+5-5=20m`
So, work done is `W=3xx20=60J`