Question

A force of $5kg-wt$ is acting on a body of mass $5kg$ kept on a smooth horizontal surface at rest. If it acts for $10$ seconds at an angle of ${60}^{\circ }$ with the horizontal, find the distance travelled by the body. (Take $g=10m/s^2)$

• A. $100m$
• B. $150m$
• C. $200m$
• D. $250m$

Answer 1

The correct option is D $250m$

$1kg-wt=gN$
Hence, force acting on body $F=5\times 10=50N.$
The component of the force along the surface $=50\cos {60}^{\circ }=25N$

Hence acceleration $a=\frac{25}{5}=5m/s^2$
Using second equation of motion, $s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 5\times {10}^2=250m$