Question

An object of mass m is moving with a constant velocity 'v'. The work done on the object to bring it to rest is

• A. mv²
• B. $\frac{1}{2}$mv²
• C. mv
• D. $\frac{1}{2}$m²v

Answer 1

The correct option is B

$\frac{1}{2}$mv²

Explanation for the correct option

B: $\frac{1}{2}$mv²

1. According to the work-energy theorem,
   When the moving body is about to stop then work is done by retarding force.
   Initial velocity of the body is v_i = v
   Final velocity of the body is v_f = 0
2. Change in kinetic energy = work done by retarding force
   Final kinetic energy - initial kinetic energy = work done by retarding force
   K.E_final - K.E_initial = W
   $\frac{1}{2}$m(v_f)² - $\frac{1}{2}$m(v_i)² = W
   where W is the work done by retarding force,
   Substituting the known values,
   $\frac{1}{2}\mathrm{m}{\left(0\right)}^2-\frac{1}{2}\mathrm{m}{\left(\mathrm{v}\right)}^2$ = W
   $-\frac{1}{2}\mathrm{m}{\left(\mathrm{v}\right)}^2$ = W
3. Work done is negative because work is done by retarding force against the motion of the body.
   Thus, the work done on the object to bring it to rest is $\frac{1}{2}$mv².
4. Hence, option (B) is correct.