Question

A particle of mass $0.1$ kg is subjected to a force which varies with distance as shown in figure. If it starts its journey from rest at $x=0$, its velocity at $x=12$ m is:

• A. $0m/s$
• B. $20\sqrt{2}m/s$
• C. $20\sqrt{3}m/s$
• D. $40m/s$

Answer 1

The correct option is D $40m/s$

As we know that area under F-x (force-displacement) graph gives work done,
hence, work=area under graph i.e., area of trapezium
=1/2×(12+4)×10 J
=1/2×16×10 J
=80 J ......(i)
Also from work-energy theorem, we know that
work done= change in kinetic energy
80=final K.E.- initial K.E. (from (i) )
80=1/2×m×$v^2$ - 0 (since body starts from rest, hence initial K.E.=0)
80=1/2×0.1×$v^2$
$v^2$=1600 (on taking square root on both sides)
v=40m/s