Question

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

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

Answer 1

The correct option is D $4.5J$


$a=\frac{F}{m}=\frac{6t}{1}=6t\frac{dv}{dt}=6t{\int }_0^{v}dv=6{\int }_0^{1}tdt\Rightarrow v=6{\left(\frac{t^2}{2}\right]}_0^1=3W=\Delta K.E=K_f-K_i=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times 3^2=4.5J$