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

Hint: 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.