Question

Velocity time graph of a particle of mass 3 Kg moving in a straight line in shown in the figure. The work done by all the forces acting on the particle in a time t = 3s is

• A. – 1800 J
• B. + 1800 J
• C. +3200 J
• D. – 3200 J

Answer 1

The correct option is A – 1800 J

Initial speed, $v_i=40m{s}^{-1}$

Final speed, $v_f=20m{s}^{-1}$

$\therefore K{E}_i=\frac{1}{2}\times 3\times {40}^2=2400J$

$K{E}_f=\frac{1}{2}\times 3\times {20}^2=600J$

$\Delta KE=600-2400=-1800J$

$\therefore$ From work – energy theorem, work done by all forces is

$W_{net}=\Delta KE=-1800J$