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Question: What is the average breaking force of a 1000-kg car moving at \(10m{{s}^{-1}}\) braking to a stop in...

What is the average breaking force of a 1000-kg car moving at 10ms−110m{{s}^{-1}} braking to a stop in 5s ?

Explanation

Solution

In the above problem, the required breaking force can be generated by friction only as there isn’t any other external force acting on the car. And this can be calculated by calculating the deceleration of the car and then multiplying it by its mass which will give us the required breaking force necessary to stop the car.

Complete step-by-step solution:
Let us first assign some terms that we are going to use later in our problem.
Let the initial velocity of the car be given by ‘u’. This initial velocity has been given to us as:
⇒u=10ms−1\Rightarrow u=10m{{s}^{-1}}
Let the time taken by the car to come to a halt be given by ‘t’. Then, the value of this time taken to stop is given as:
⇒t=5s\Rightarrow t=5s
Also, let the final velocity of the car be given by ‘v’ and its deceleration (assumed to be constant) be given by ‘a’.
Then, using equation of motion of a body moving in a straight line, we get:
⇒v=u+at\Rightarrow v=u+at [Let this expression be termed as equation number (1)]
Putting the values of all the respective terms in the above equation and calculating for the deceleration of car, we get:
⇒0=10+a(5) ⇒a=−2ms−2 \begin{aligned} & \Rightarrow 0=10+a(5) \\\ & \Rightarrow a=-2m{{s}^{-2}} \\\ \end{aligned}
Hence, the braking force applied on the car is equal to:
⇒∣F∣=ma\Rightarrow \left| F \right|=ma
Where, ‘m’ is the mass of the car and is equal to 1000kg.
Thus, the magnitude of braking force is equal to:
⇒∣F∣=1000×2N ∴∣F∣=2000N \begin{aligned} & \Rightarrow \left| F \right|=1000\times 2N \\\ & \therefore \left| F \right|=2000N \\\ \end{aligned}
Hence, the average breaking force of a 1000-kg car moving at 10ms−110m{{s}^{-1}} braking to a stop in 5s comes out to be 2000N.

Note: In the above solution, we could see that the sign of acceleration comes out to be negative. This implies that it is acting in a direction opposite to the initial velocity direction (which was taken as positive in the solution). This also proves that we are moving head in the right direction in terms of solving the problem correctly.