Equation for Work required to achieve a certain velocity [closed]

Question

If we are trying to find the work required to get an object moving at velocity $v$, and we start with $w = f\cdot d$, we can then make the following substitutions:

substitute $f$ with ma: $w = m\cdot a\cdot d$

substitute $d$ with $(\frac{1}{2})at^2: w = (\frac{1}{2})m \cdot a \cdot at^2$

substitute $t$ with $\frac{v}{a}$: $w = (\frac{1}{2})m v^2$

At the start, $w$ is equal to $f\cdot d$ which is also equal to $mv^2$, yet we conclude by saying that w is also equal to $(\frac{1}{2})mv^2$. I am not sure what is causing the contradiction.

Answer 1

This contradiction arose since you took a distance to be = $v.t$ which won't be the right formula to use here since the body is accelerating. Thus you got $mv^2$.

Answer 2

Your derivation above is valid for a constant force $F$ acting on an object that starts from rest.

By the definition of net work $W_{\rm net}$, we have that the net work equals the sum of all the work done by all the forces acting on the object. $$W_{\rm net}\equiv \sum_i W_i = \Delta {\rm KE} = \dfrac1 2 mv_f^2-\dfrac1 2 mv_i^2$$

For the particular 1-dimensional case that you have a constant force that does work $W=Fd$ and the object starts from rest, you have

$$W_{\rm net}= Fd \implies Fd = \dfrac1 2 mv_f^2$$

As from looking over your work, I do not believe that the contradiction of the work being $mv^2$ ever arises, unless you made a mistake on paper which you did not reproduce here. Nevertheless, the net work equation I wrote out above always holds, without any contradictions. So you can be sure that the sum of all works will always equal the change in kinetic energy.

EDIT: JustJohan's answer explains why you probably thought the work was $mv^2$.