Question

When a force act on a body of mass its position x varies with time t as x=at^4+bt+c where a,b,c areconstants workdone is

Answer 1

Question :-

When a constant force F acts on a body of mass m, its position x varies with time t as $\sf{x=at^4+bt+c}$ where a, b, c are constants. Find the work done by the force F.

Answer:-

$\Large\boxed{\sf{\quad W=Ft(at^3+b)+k\quad}}$

Solution:-

In case of varying force / displacement, we have, work done,

$\displaystyle\longrightarrow\sf{W=\int F\ dx}$

Here, the position of the body, x, varies with time t as,

$\displaystyle\longrightarrow\sf{x=at^4+bt+c}$

Differentiating wrt time,

$\displaystyle\longrightarrow\sf{\dfrac{dx}{dt}=\dfrac{d}{dt}\left[at^4+bt+c\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{dx}{dt}=4at^3+b}$

$\displaystyle\longrightarrow\sf{dx=(4at^3+b)\ dt}$

Hence the work done is,

$\displaystyle\longrightarrow\sf{W=\int F\ dx}$

$\displaystyle\longrightarrow\sf{W=\int F\ (4at^3+b)\ dt}$

$\displaystyle\longrightarrow\sf{W=\int(4Fat^3+Fb)\ dt}$

$\displaystyle\longrightarrow\sf{W=\int4Fat^3\ dt+\int Fb\ dt}$

$\displaystyle\longrightarrow\sf{W=4Fa\int t^3\ dt+Fb\int dt}$

$\displaystyle\longrightarrow\sf{W=4Fa\cdot\dfrac{t^4}{4}+Fb\cdot t+k}$

$\displaystyle\longrightarrow\sf{\underline{\underline{W=Fat^4+Fbt+k}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{W=Ft(at^3+b)+k}}}$

where 'k' is the integral constant.