Question

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 :

• A. 4.5 J
• B. 22 J
• C. 9 J
• D. 18 J

Answer 1

Answer: (A)

4.5 J

Answer 2

$6 t=1 \cdot \frac{d v}{d t}$
\int_\limits{0}^{v} d v=\int 6t\, d t
$v=6\left[\frac{t^{2}}{2}\right]_{0}^{1}$
$=3\, ms ^{-1}$
$W=\Delta KE =\frac{1}{2} \times 1 \times 9=4.5\, J$