Explain with graphs the difference between work done by a constant force and by a variable force.

Explanation

Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
`dW (F costheta)dr` ...(1)
The total work done in producing a displacement from initial position `r_(i)` to final position `r_(f)` is
`W=int_(r_(i))^(r_(f))dW` ...(2)
`W=int_(r_(i))^(r_(f))(F costheta)dr=(F costheta)int_(r_(i))^(r_(f))dr=(F costheta)(r_(f)-r_(i))` ...(3)
The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force: When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation

`dW=F costheta dr` [F `costheta` is the component of the variable force F]
where, F and `theta` are variables. The total work done for a displacement from initial position `r_(i)`mentfinal position `r_(f)` is given by the relation,
` W=int_(r_(i))^(r_(f))dW=int_(r_(i))^(r_(f))F costheta dr` ...(4)
A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.
Expression of power and velocity
The work done by force `vecF` for a displacement `dvecr` is
`W=intvecF.dvecr` ...(1)
Left hand side of the equation (1) can be written as
`W=intdW=int(dW)/(dt)dt` (multiplied and divided by dt) ...(2)
Since, velocity is `vecv=(dvecr)/(dt)`,`dvecr=vecvdt`. Right hand side of the equation (1) can be written as
`intvecF.dvecr=int(vecF.(dvecr)/(dt))dt=int(vecF.vecv)dt[vecv=(dvecr)/(dt)]` ...(3)
Substituting equation (2) and equation (3) in equation (1), we get
`int(dW)/(dt)dt=int(vecF.vecv)dt`
`int((dW)/(dt)-vecF.vecv)dt=0`
This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,
`(dW)/(dt)-vecF.vecv=0` Or `(dW)/(dt)=vecF.vecv`
Hence power `P=vecF.vecv`