Question

On applying the brakes, a car of 1000 kg moving at $36\frac{km}{h}$ stops. Find the total work done on the car.

• A. $-5\times {10}^{3}J$
• B. $5\times {10}^{3}J$
• C. $-5\times {10}^{4}J$
• D. $5\times {10}^{4}J$

Answer 1

The correct option is C $-5\times {10}^{4}J$

Initial speed, $u=36\frac{km}{h}=\frac{36\times 1000}{3600}\frac{m}{s}=10\frac{m}{s}$
Initial kinetic energy, $E_i=\frac{1}{2}m{u}^2$
Final kinetic energy, $E_f=0$
Change in kinetic energy, $\Delta E=E_f-E_i=-\frac{1}{2}m{u}^2$
Using work-energy theorem, work done on the car equals to change in its kinetic energy.
$W=\Delta E=-\frac{1}{2}m{u}^2$
$W=-\frac{1}{2}\times 1000\times {10}^2=-5\times {10}^{4}J$
Note that work done is found negative. This means that force and displacement were in opposite directions.