An object of mass $100 \mathrm{kg}$ is accelerated uniformly from a velocity of $5 \mathrm{m}/\mathrm{s}$ to $8 \mathrm{m}/\mathrm{s}$ in $6 \mathrm{s}.$ Calculate the magnitude of the force exerted on the object?

The magnitude of the force exerted on the object is $50\mathit{ N}$
Given,
Mass of the object, $\mathrm{ }=\mathrm{ }100 \mathrm{kg}$, 
initial velocity, $\mathrm{u} =\mathrm{ }5 \mathrm{m}/\mathrm{s}$
Final velocity, $\mathrm{ }=\mathrm{ }8 \mathrm{m}/\mathrm{s}$, 
time taken, $\mathrm{t}=\mathrm{ }6 \mathrm{s}$
Hence,
Initial momentum ${\mathrm{p}}_1= \mathrm{m}\times \mathrm{u}$$=100\mathrm{ }\times \mathrm{ }5$                               

${\mathrm{p}}_1=500 \mathrm{kg}m/\mathrm{s}$.
Final momentum ${\mathrm{p}}_2= \mathrm{m}\times \mathrm{v}$

${\mathrm{p}}_2=100\times 8$             

$=800 \mathrm{kg}m/\mathrm{s}$ 

By the second law of motion, the magnitude of the applied unbalanced force equals the rate of change of momentum. Mathematically,
$\mathrm{F}=\frac{\mathrm{\Delta p}}{\mathrm{\Delta t} }\mathrm{ }$
$\Rightarrow \mathrm{F}=\frac{{\mathrm{p}}_2\mathrm{ }-\mathrm{ }{\mathrm{p}}_1}{\mathrm{t}}\mathrm{ }$
          $=\frac{ ( 800 - 500 )}{6}$
          $=\frac{300}{6}\mathit{ }$
          $=\mathrm{ }50 \mathrm{kg}m/\mathrm{s}$  

$\Rightarrow \mathrm{F}=50 \mathrm{N}$