Question

A force of $5\text{kg – wt}$ is acting on a body of mass $5\text{kg}$ 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 it along the surface (Take $g=10{\text{m/s}}^2$)

• A. $100\text{m}$
• B. $150\text{m}$
• C. $200\text{m}$
• D. $250\text{m}$

Answer 1

The correct option is D $250\text{m}$

We know that
$1\text{Kg - wt=g N}=10\text{N}$
Hence, $F=5\times 10\text{N}=50\text{N}$
The component of the force along the horizontal surface $=F_s=50\cos {60}^{\circ }=25\text{N}$
Hence, acceleration of the body along the surface, $a=\frac{F_s}{m}=\frac{25}{5}=5{\text{m/s}}^2$
Distance travelled by the body along the surface in $10\text{s}$,
$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 5\times 100=250\text{m}$