Question

The work done required to stop a car of mass $100kg$ moving at a velocity of $20m{s}^{-1}$ will be:-

• A. $2\times {10}^{5}J$
• B. $50J$
• C. $-50J$
• D. $-2\times {10}^{4}J$

Answer 1

The correct option is D $-2\times {10}^{4}J$

Work done by an external force is the change in kinetic energy of the car.

$\Delta K=(KE{)}_{final}-(KE{)}_{initial}$

Finally, the car will stop so it's kinetic energy will be zero.

$(KE{)}_{initial}=\frac{1}{2}m{v}^2$

$m=100kg$

$v=20m{s}^{-1}$

$\Rightarrow (KE{)}_{initial}=\frac{1}{2}\times 100\times {20}^2$

$\Rightarrow (KE{)}_{initial}=2\times {10}^{4}J$

$\Rightarrow \text{work done}=0-(2\times {10}^4)J$

$\Rightarrow \text{work done}=-(2\times {10}^4)J$