Question

A truck weighing 1000kgf changes its speed from $36\,km\,h{{r}^{-1}}$ to $72km\,h{{r}^{-1}}$ in $2\,\min $. Calculate:
(a). work done by the engine
(b). its power
($g=10m{{s}^{-1}}$)

Answer 1

Hint: A gravitational pull is acting on the truck and its magnitude is given in kilogram-force. Kilogram-force is a unit of force, using force we can calculate the acceleration and mass of the body and hence the work done on it by the engine. Power is the rate of doing work.

Formulas used:
$a=\dfrac{v-u}{t}$
$F=ma$
$W={{E}_{2}}-{{E}_{1}}$
$P=\dfrac{W}{t}$

Complete step by step solution:
Converting the units to SI for velocity-
$36\,km\,h{{r}^{-1}}=\dfrac{36\times 1000}{1\times 3600}m{{s}^{-1}}$
$\Rightarrow 36\,km\,h{{r}^{-1}}=10m{{s}^{-1}}$ ------------- (1)
$72km\,h{{r}^{-1}}=\dfrac{72\times 1000}{1\times 3600}m{{s}^{-1}}$
$\Rightarrow 72km\,h{{r}^{-1}}=20m{{s}^{-1}}$ ----------- (2)
The acceleration is the change in velocity per unit time; therefore we substitute the values of given initial and final velocities to calculate acceleration,
$\begin{align}
  & a=\dfrac{v-u}{t} \\
 & \Rightarrow a=\dfrac{20-10}{2\times 60}m{{s}^{-2}} \\
 & \therefore a=\dfrac{1}{12}m{{s}^{-2}} \\
\end{align}$
$kgf$ is the gravitational unit of metric system,$1\,kgf$ is equal to
$\begin{align}
  & 1\,kgf=1\times 10N \\
 & \therefore 1\,kgf=10N \\
\end{align}$
 So, the $1000\,kgf$will be-
$\begin{align}
  & 1000\,kgf=10\times {{10}^{3}}N \\
 & \therefore 1000\,kgf={{10}^{4}}N \\
\end{align}$
We know that force is the product of mass and acceleration, therefore,
$F=ma$
Here,$F$ is the force
$m$ is the mass of the body
$a$ is the acceleration
We substitute values in the above equation to get,
$\begin{align}
  & {{10}^{4}}=m\times \dfrac{1}{12} \\
 & \Rightarrow 12\times {{10}^{4}}=m \\
 & \therefore m=12\times {{10}^{4}}kg \\
\end{align}$
The initial kinetic energy of the truck is-
$\begin{align}
  & {{E}_{1}}=\dfrac{1}{2}mv_{1}^{2} \\
 & \Rightarrow {{E}_{1}}=\dfrac{1}{2}\times 12\times {{10}^{4}}\times {{(10)}^{2}} \\
 & \therefore {{E}_{1}}=6\times {{10}^{6}}J \\
\end{align}$
The final kinetic energy of the truck is-
$\begin{align}
  & {{E}_{2}}=\dfrac{1}{2}mv_{2}^{2} \\
 & \Rightarrow {{E}_{2}}=\dfrac{1}{2}\times 12\times {{10}^{4}}\times {{(20)}^{2}} \\
 & \therefore {{E}_{2}}=24\times {{10}^{6}} \\
\end{align}$
Work done is the change in energy of a body, therefore the change in energy of the truck is-
$\begin{align}
  & W={{E}_{2}}-{{E}_{1}} \\
 & \Rightarrow W=(24-6)\times {{10}^{6}} \\
 & \therefore W=18\times {{10}^{3}}kJ \\
\end{align}$
Therefore, the work done by the truck is $18\times {{10}^{3}}kJ$

Power is rate of doing work, it is given by-
$P=\dfrac{W}{t}$
Therefore, we substitute values in the above equation to get,
$\begin{align}
  & P=\dfrac{18\times {{10}^{6}}}{2\times 60} \\
 & \Rightarrow P=150\times {{10}^{3}}W \\
 & \therefore P=150kW \\
\end{align}$

Therefore, the power of the engine is $150kW$.

Note:
The system of the truck is not isolated as it experiences gravitational force. According to Newton’s second law, a force is required to change the state of rest or motion of an object. The work done is a scalar quantity. It is the scalar product of force vector and velocity vector. Work done is also calculated as the product of force and displacement.