Question

A time dependent force F t  6 acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. Will be

Answer 1

Correct Question:

A time dependent force F = 6 t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec will be?

Answer:

• The work done (w) by the force is 4.5 Joules.

Given:

1. Relation :- F = 6 t.
2. Mass of particle (m) = 1 Kg
3. Time period = 1 second.

Explanation:

$\rule{300}{1.5}$

From the relation we Know,

$\large\bigstar \; \boxed{\tt F = Ma}$

$\mathfrak{Here}\begin{cases}\text{F Denotes Force}\\ \text{M Denotes Mass of Body}\\\text{a Denotes acceleration}\end{cases}$

Now,

$\large \boxed{\tt F = Ma}$

Substituting the values,

$\displaystyle \dashrightarrow\tt F=m.\dfrac{dv}{dt} \ \ \ \because\left(a=\dfrac{dv}{dt}\right)\\\\\\\dashrightarrow\tt F=1\times\dfrac{dv}{dt}\ \ \ \because(m=1\;Kg)\\\\\\\dashrightarrow\tt 6\;t=\dfrac{dv}{dt}\ \ \ \because(F=6\;t)\\\\\\\dashrightarrow\tt dv=6\;t.dt$

Integrating, and applying limits.

$\displaystyle \dashrightarrow\tt \displaystyle\int\limits^{v}_{0}\tt dv=\displaystyle\int\limits^{1}_{0}\tt 6\;t.dt\\\\\\\dashrightarrow\tt \Bigg[v\Bigg]^{v}_{0}=\Bigg[\dfrac{6\;t^2}{2}\Bigg]^{1}_{0}\\\\\\\dashrightarrow\tt (v-0)=\Bigg[3\;t^2\Bigg]^{1}_{0}\\\\\\\dashrightarrow\tt v=3\;([1]^2-[0]^2)\\\\\\\dashrightarrow \large{\underline{\blue{\tt v=3\;m/s}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From work Energy theorem.

$\large\bigstar \; \boxed{\tt W = \Delta K.E}$

$\mathfrak{Here}\begin{cases}\text{W Denotes Work done}\\ \sf{\Delta}\text{ K.E Denotes Change in K.E}\\\end{cases}$

Now,

$\large \boxed{\tt W = \Delta K.E}$

Substituting the values,

$\displaystyle \dashrightarrow\tt W=\dfrac{1}{2}\;m\;(v^2-u^2)\\\\\\\dashrightarrow\tt W=\dfrac{1}{2}\times 1 \times (3^2-0^2)\ \because[u=10\;m/s\;\&\;v=3\;m/s]\\\\\\\dashrightarrow\tt W=\dfrac{1}{2}\times 1 \times 9\\\\\\\dashrightarrow\tt W=\dfrac{9}{2}\\\\\\\dashrightarrow\tt W=\cancel{\dfrac{9}{2}}\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt W=4.5\;J}}}}$

∴ The work done (w) by the force is 4.5 Joules.

$\rule{300}{1.5}$

Answer 2

$\huge\bold\green{Question}$

A time dependent force F = 6 t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. Will be

$\huge\bold\green{Answer}$

According to the question we have given :-

$\begin{lgathered}\tt{Given}\begin{cases}\text{Relation : F = 6 t}\\ \text{Mass of partical = 1 kg}\\\text{ Time taken = 1 sec .}\end{cases}\end{lgathered}$

So , as we know that the relation :-

$\huge\sf\orange{F = Ma}$

• F = Force
• M = Mass of Body
• a = aacceleratio

Now , by Substituting the known values in this relation :-

$\begin{lgathered}\displaystyle \sf =F=m.\dfrac{dv}{dt} \\ \\ = \left(a=\dfrac{dv}{dt}\right)\\\\\\\sf =F=1\times\dfrac{dv}{dt}\ \ \ \(m=1\;Kg)\\\\\\\sf 6\;t=\dfrac{dv}{dt}\\ \\\because(F=6\;t)\\\\\\\sf =dv=6\;t.dt\end{lgathered}$

Now by Integrating this , and applying limits on above values we get

$\begin{lgathered}\displaystyle \sf \displaystyle\int\limits^{v}_{0}\tt dv=\displaystyle\int\limits^{1}_{0}\tt 6\;t.dt\\\\\\\sf= \Bigg[v\Bigg]^{v}_{0}=\Bigg[\dfrac{6\;t^2}{2}\Bigg]^{1}_{0}\\\\\\\sf= (v-0)=\Bigg[3\;t^2\Bigg]^{1}_{0}\\\\\\= v=3\;([1]^2-[0]^2)\\\\\\end{lgathered}$

Now by using the work Energy theorem :-

$\huge\sf\orange {W = \Delta K.E}$

• W = Work done
• Δ K.E = Change in K.E

Now , by substituting the known values in this relation

$\begin{lgathered}\displaystyle \sf= W=\dfrac{1}{2}\;m\;(v^2-u^2)\\\\\\\sf= W=\dfrac{1}{2}\times 1 \times (3^2-0^2)\ \because[u=10\;m/s\;\&\;v=3\;m/s]\\\\\\\sf= W=\dfrac{1}{2}\times 1 \times 9\\\\\\\sf= W=\dfrac{9}{2}\\\\\\\sf= W=\cancel{\dfrac{9}{2}}\\\\\\\sf ={ W=4.5\;J}\end{lgathered}$

Hence , the work done (w) by the force are 4.5 Joules.