Question

A truck weighing 1000kgf changes its speed from $36km{{h}^{-1}}$ to $72km{{h}^{-1}}$in 2 minutes. Calculate its power ($g=10m{{s}^{-1}}$).
$\text{A}\text{. }1.25W$
$\text{B}\text{. }1.25\times {{10}^{6}}W$
$\text{C}\text{. }1.25\times {{10}^{3}}W$
$\text{D}\text{. }1.25\times {{10}^{4}}W$

Answer 1

Power is the rate of work done or energy consumed or dissipated with respect to time. It tells us the work done or energy consumed or dissipated in one unit of time. Kinetic energy is one form of energy. Therefore, the amount of change in kinetic energy per unit time is power.

Formula used:
Kinetic energy: - $K.E=\dfrac{1}{2}mv_{2}^{2}$
Power: $P=\dfrac{\Delta E}{t}$

Complete step by step answer:
When a body of mass m is in a motion with speed v, it possesses an energy called kinetic energy (K.E). The kinetic energy of the body is given as ${{v}_{1}}$. …..(1)
The SI unit of energy is joules (J). Since the kinetic energy of a body is directly proportional to the square of the speed of the body, when the speed of the body changes, its kinetic energy also changes.
When a body changes its speed from ${{v}_{1}}$ to ${{v}_{2}}$ its kinetic energy will change from $K.{{E}_{1}}$ to $K.{{E}_{1}}$. Therefore, the change in kinetic energy, denoted as $\Delta K.E$, will be $\Delta K.E=K.{{E}_{2}}-K.{{E}_{1}}$. According to equation (1), the kinetic energy at speed ${{v}_{1}}$ will be equal to $\dfrac{1}{2}mv_{1}^{2}$ and the kinetic energy at speed ${{v}_{2}}$ will be equal to $\dfrac{1}{2}mv_{2}^{2}$. Therefore, change in kinetic energy will be
$\Delta K.E=K.{{E}_{2}}-K.{{E}_{1}}=\dfrac{1}{2}mv_{2}^{2}-\dfrac{1}{2}mv_{1}^{2}=\dfrac{1}{2}m\left( v_{2}^{2}-v_{1}^{2} \right)$.

The rate of change in energy with time is called power. It tells us the amount of energy changing in one second. Average power (P) is the total change in energy ($\Delta E$) upon the given time (t) for which it changes. Therefore, $P=\dfrac{\Delta E}{t}$.

With this theory, we can now solve the given question. To calculate the power, we need to find the change in energy (kinetic energy). Before that, let us write the given speeds in SI units i.e. $m{{s}^{-1}}$.
Therefore,

$36km{{h}^{-1}}=36\times (1000m)\times {{(60\times 60\operatorname{s})}^{-1}}=\dfrac{36000m}{3600s}=10m{{s}^{-1}}$.
Since 72 is two times 36, $72km{{h}^{-1}}=20m{{s}^{-1}}$.
So the speed of the truck changes from $10m{{s}^{-1}}$ to $20m{{s}^{-1}}$.
Therefore, ${{v}_{1}}=10m{{s}^{-1}}$ and ${{v}_{2}}=20m{{s}^{-1}}$.

Hence, the change in kinetic energy will be $\Delta K.E=\dfrac{1}{2}m\left( v_{2}^{2}-v_{1}^{2} \right)=\dfrac{1}{2}\times (1000)\times \left( {{20}^{2}}-{{10}^{2}} \right)=500\times \left( 400-100 \right)=150000J$

This change in the kinetic energy of the truck takes place for 2minutes. Therefore, its power is $P=\dfrac{\Delta E}{t}=\dfrac{150000J}{2\min }=\dfrac{150000J}{2\times 60s}=1250W=1.25\times {{10}^{3}}W$
Hence, the correct option is (C).

Note:
It has to be noted that the v in the formula of the kinetic energy is speed and not velocity. Sometimes, the velocity of a body may be changing but if its speed is constant then it means its kinetic energy is also constant. For example, a body in uniform circular motion has changing velocity but the kinetic energy of the body is constant due to the constant speed.