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Q.
A time dependent force $F = 6t$ 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 :
Given,
$
F =6 t
$
$
m =1 kg
$
Force, $F = ma =6 t$
$
\begin{array}{l}
1 \times \frac{d v}{d t}=6 t \\
d v=6 t d t
\end{array}
$
Integrating both side,
$
\begin{array}{l}
v =\int dv =\int_0^1 6 tdt \\
v =6\left[\frac{ t ^2}{2}\right]_0^1 \\
v =3 m / s
\end{array}
$
Initial velocity, $u =0 m / s$
From work energy theorem,
$
\begin{aligned}
W &= K _{ f }- K _{ i } \\
W &=\frac{1}{2} mv ^2-\frac{1}{2} mu ^2 \\
W &=\frac{1}{2} \times 1 \times 3 \times 3-\frac{1}{2} \times 1 \times 0 \times 0 \\
W &=4.5-0=4.5 J
\end{aligned}
$