Question

A car moving at $54km{h}^{-1}$is to be stopped by applying brakes in the next 5m If the car weights $1800kgwt$ then the net external force acting on it is

• A. $4\times {10}^{9}N$
• B. $40N$
• C. $4\times {10}^{4}dyne$
• D. $4\times {10}^{4}N$

Answer 1

The correct option is D

$4\times {10}^{4}N$

Step 1: Given data

The initial velocity of the car u =$54km{h}^{-1}$

Converting the value of initial velocity to ms^-1

$54km{h}^{-1}=\frac{54\times 1000}{60\times 60}=15m{s}^{-1}$

Distance traveled s =$5m$

The final velocity of the car v= 0 ( because the car finally stops after applying brakes)

Weight of the car W=$1800kgwt$

Step 2: Calculations

The kinematic equation gives the relation between the physical quantities distance, velocity acceleration, and time.

From the third kinematic equation we have,

$v^2=U^2+2as$

Where V is the final velocity, U is the initial velocity, a is the acceleration and s is the displacement.

Substituting the given values into the kinematic equation,

$0={15}^2+2as$

Rearranging this equation to final acceleration(a),

$$\begin{gathered} a=\frac{-{15}^2}{2\times 5} \\ a=-22.5m{s}^{-2} \end{gathered}$$

So, the acceleration of the car is $-22.5m{s}^{-2}$

The negative sign indicate that the acceleration is decreasing with an increase in time.ie; retardation

From Newton's formula

$$\begin{gathered} F=ma \\ Substitutingthevalueofa \\ F=1800\times 22.5 \\ F=4\times {10}^{4}N \end{gathered}$$

The net external force acting on the car is $4\times {10}^{4}N$

Hence, option (d) is correct