Question

A car weighing $500kg$.working against the resistance of $500N$ accelerates from rest to $20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^{-1}$}\right.$ in $100m$.$\left(g=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right).$The work done by the engine of the car is

• A. $1.0\times {10}^{5}J$
• B. $1.5\times {10}^{5}J$
• C. $1.05\times {10}^{5}J$
• D. The information is given insufficient

Answer 1

The correct option is B

$1.5\times {10}^{5}J$

Step 1: According to the question we have given that

Mass of car $m=500kg$

Initial velocity of a car$u=0$

Final velocity of a car $v=20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^{-1}$}\right.$

Step 2: Calculate the acceleration of the body,

From the kinematics equation,

$$\begin{gathered} \Rightarrow v^2=u^2+2as \\ \Rightarrow {\left(20\right)}^2=0+2\times a\times 100 \\ \Rightarrow 400=200\times a \\ \Rightarrow a=2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Step 3: Calculate the work done by the engine,

According to Newton's second law,

$$\begin{gathered} \Rightarrow F_{net}=m\times a \\ \Rightarrow F_{net}=500\times 2 \\ \Rightarrow F_{net}=1000N \end{gathered}$$

Force by engine,

$$\begin{gathered} \Rightarrow F-500=1000 \\ \Rightarrow F=1500N \end{gathered}$$

Work done by the engine of car,

$$\begin{gathered} \Rightarrow W=F\times D \\ \Rightarrow W=1500\times 100 \\ \Rightarrow W=150000 \\ \Rightarrow W=1.5\times {10}^{5}J \end{gathered}$$

Thus, the work done by the engine is $1.5\times {10}^{5}J$.

Hence, option B is the correct answer.