# A body of mass 10 kg at rest is acted upon simultaneously by two forces 4N and 3N at right angles to each other. The kinetic energy of body at end of 10sec is

$(a)\;100 \;J \quad (b)\;300\;J\quad (c)\;50\;J \quad (d)\;125\; J$

Since it is acted upon simultaneously by two forces 4N and 3N at right angles, net force $=\sqrt {4^2+3^2} = 5N$
The mass is given as $10\;Kg \rightarrow \;a=\large\frac{F}{m}$$=\large\frac{5}{10}$$ = 0.5\;m/s^2$
Now, we can calculate the Kinetic Energy as $K.E=\large\frac{1}{2} $$mv^2=\large\frac{1}{2}$$m(at)^2$
$\Rightarrow KE =\large\frac{1}{2}$$10 \times 0.5^2 \times 10 s$$ = 125 J$
edited Aug 25, 2014