Question

The work required to be done to stop an object of 300 kg moving at a velocity of $36km/h$

• A. $10kJ$
• B. $-7kJ$
• C. $-15kJ$
• D. $-10kJ$

Answer 1

The correct option is C $-15kJ$

Mass of the object, m = 300 kg
Initial velocity, $u=36km/h=\frac{36\times 5}{18}=10m/s$

Final velocity, v = 0

According to work energy theorem:

Work done, W = Change in kinetic energy of the object

$=(K.E{)}_f-(K.E{)}_f=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2\Rightarrow W=\frac{1}{2}m(v^2-u^2)\Rightarrow W=\frac{1}{2}\times 300[0-(10{)}^2]\Rightarrow W=150\times (-100)=-15000J\Rightarrow =-15kJ$

Hence, the correct answer is option (c).