Question

Calculate the work required to be done to stop a car of $1500\,{\text{kg}}$ moving at a velocity of $60\,{\text{km/h}}$.

Answer 1

Use the formula for kinetic energy and determine the initial and final kinetic energy of the car. Then use the work-energy theorem which gives the relation between the change in kinetic energy of the car with the work done on the car..

Formula used:
The kinetic energy $K$ of an object is given by
$K = \dfrac{1}{2}m{v^2}$ …… (1)
Here, $m$is the mass of the object and $v$ is the velocity of the object.
The work-energy theorem is
$W = \Delta K$ …… (2)
Here, $W$is the work done and $\Delta K$ is the change in the kinetic energy.

Complete step by step answer:
The mass of the car is $1500\,{\text{kg}}$ and velocity is $60\,{\text{km/h}}$.
$m = 1500\,{\text{kg}}$
$v = 60\,{\text{km/h}}$

Convert the unit of velocity in the SI system of units.
$v = \left( {60\,\dfrac{{{\text{km}}}}{{\text{h}}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{h}}}}{{3600\,{\text{s}}}}} \right)$
$\Rightarrow v = 16.66\,{\text{m/s}}$
Hence, the velocity of the car is $16.66\,{\text{m/s}}$.
Calculate the initial kinetic energy ${K_i}$ of the car.
${K_i} = \dfrac{1}{2}m{v^2}$
Substitute $1500\,{\text{kg}}$ for $m$ and $16.66\,{\text{m/s}}$ for $v$ in the above equation.
${K_i} = \dfrac{1}{2}\left( {1500\,{\text{kg}}} \right){\left( {16.66\,{\text{m/s}}} \right)^2}$
$\Rightarrow {K_i} = 208166.7\,{\text{J}}$
Hence, the initial kinetic energy of the car is $208166.7\,{\text{J}}$.
Since the car stops when the force is applied, the final kinetic energy ${K_f}$ of the car becomes zero.
${K_f} = 0\,{\text{J}}$
Calculate the change in kinetic energy $\Delta K$ of the car.
$\Delta K = {K_f} - {K_i}$
Substitute $0\,{\text{J}}$ for ${K_f}$ and $208166.7\,{\text{J}}$ for ${K_i}$ in the above equation.
$\Delta K = \left( {0\,{\text{J}}} \right) - \left( {208166.7\,{\text{J}}} \right)$
$\Rightarrow \Delta K = - 208166.7\,{\text{J}}$
$\Rightarrow \Delta K = - 208.17\,{\text{kJ}}$
Hence, the change in kinetic energy of the object is $- 208.17\,{\text{kJ}}$.
Calculate the work done to stop the car.
Substitute $- 208.17\,{\text{kJ}}$ for $\Delta K$ in equation (2).
$W = - 208.17\,{\text{kJ}}$.

Hence, the work done to stop the car is $208.17\,{\text{kJ}}$.

Note:
The negative sign indicates that the work done indicates that the force is applied in a direction opposite as that of the car.
Also in this above given solution the car stops when the force is applied, the final kinetic energy ${K_f}$ of the car becomes zero.