Question

Under a force an object of mass 2 kg moves
such that x =
${t}^{3} \div 3$
where x is in metre and
t is in second. Work done in first 2 sec is :-
(A) 1.6 J
(B) 16 J
(C) 160 J
(D) 1600 J​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{ x = \frac{t^3}{3}}$

Mass (m) = 2 Kg.

$\huge{\bold{\underline{Explanation:-}}}$

Firstly finding the velocity,

As we know,

$\large{v = \frac{dx}{dt}}$

Substituting the value

$\large{v = \frac{d(t^3/3)}{dt}}$

Now, Differentiating

$\large{v = 3 \times \frac{t^2}{3}}$

$\large{v = \cancel{3} \times \frac{t^2}{ \cancel{3}}}$

it becomes,

$\large{\boxed{v = t^2}}$

From Work - energy theorem,

$\large{\bold{W = \frac{1}{2} mv^2 - \frac{1}{2}mu^2}}$

As the velocity is depending on time,

the initial velocity at (t = 0 second) will be also Zero.

Now,

$\large{W = \frac{1}{2}mv^2 - 0}$

$\large{W = \frac{1}{2}mv^2}$

Substituting the values,

$\large{W = \frac{1}{2} \times 2 \times (t^2)^2}$

Using identity,

$\large{(a^n)^m = a ^{nm}}$

As t = 2 seconds.

$\large{W = \frac{1}{2} \times 2 \times (2)^4}$

$\large{W = \frac{1}{ \cancel{2}} \times \cancel{2} \times 16}$

$\huge{\boxed{\boxed{W = 16 \: J}}}$

So, the work done in first 2 seconds is 16 Joules (Option --B).