Question

A car moving at $40km{h}^{-1}$is to be stopped by applying brakes in the next $4m$ If the average resultant force on it is $3.1\times {10}^{4}N$then the mass of the car is _____ kg.

• A. $2009$
• B. $2500$
• C. $1500$
• D. $1600$

Answer 1

The correct option is A

$2009$

Step 1: Given data

The initial velocity of the car U=$40km{h}^{-1}$

The final velocity of the car V= 0 km h^-1

Distance traveled after applying brakes S=$4m$

Force F= $3.1\times {10}^{4}N$

Step 2: Calculations

From the kinematic equation,

$$\begin{gathered} V^2=U^2+2aS \\ a=\frac{V^2-U^2}{2S} \\ a=\frac{0-\left({\displaystyle \frac{40\times 5}{18}}\right)}{2\times 4}=\frac{-200\times 200}{18\times 18\times 8} \\ a=\frac{-40000}{2592}=\frac{-4\times {10}^4}{2592}m{s}^{-2} \end{gathered}$$

From Newton's law, we have

$$\begin{gathered} F=ma \\ m=\frac{F}{a}=\frac{3.1\times {10}^4}{{\displaystyle \raisebox{1ex}{$-4\times {10}^4$}\!\left/ \!\raisebox{-1ex}{$2592$}\right.}} \\ m=3.1\times {10}^4\times \frac{2592}{-4\times {10}^4} \\ m=2008.8kg \\ m=2009kg \end{gathered}$$

Hence,

Option (a) is correct