Question

A body of mass m is moving with a uniform velocity u.A force is applied on the body due to which its velocity increases from u to v.How much work is being done by the force?

Answer 1

Hey Pretty Stranger!

Let the work done be w, force be f, displacement be s and acceleration be a

Using 3rd eqn of motion :

★ v² = u² + 2as

$\therefore \sf \: a = \dfrac{ {v}^{2} - {u}^{2} }{2s}$

We know that Force = Mass × Acceleration

$\longrightarrow \sf \: Force = \dfrac{ m( {v}^{2} - {u}^{2}) }{2s}$

Also, Work done = Force × Displacement

Thus,

$\longrightarrow \sf \: W = \dfrac{ m( {v}^{2} - {u}^{2})s }{2s}$

$\therefore\large\boxed{\sf W\: = \dfrac{1}{2} m({v}^{2} \: - {u}^{2})}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Work\:Done=\frac{m(v^{2}-u^{2})}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: body = m \\ \\ \tt: \implies Initial \: velocity = u \\ \\ \tt: \implies Final \: velocity = v \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Work \: Done =?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies F= ma - - - - - (1) \\ \\ \bold{For \: work \: done : } \\ \tt: \implies Work \: Done = Force \times Displacement \\ \\ \tt: \implies Work \: Done = ma \times s \\ \\ \tt \circ \: s = \frac{ {v}^{2} - {u}^{2} }{2a} - - - - - (2) \\ \\ \tt: \implies Work \: Done = m \times a \times \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \green{\tt: \implies Work \: Done = \frac{m( {v}^{2} - {u}^{2} )}{2} }$