Question

How much force is required to bring a car of mass 1,273 kg, initially traveling at 28 m/s, to rest in 5.8 s?

Answer 1

We are asked to determine the force required to stop a car traveling at a speed of 28 m/s in 5.8s. To do that we will calculate the acceleration of the car first. The acceleration is determined using the following equation:

`v_f=v_0+at_{}`

Where

`\begin{gathered} v_f,v_0=\text{ final and initial velocities} \\ a=\text{ acceleration} \\ t=\text{ time} \end{gathered}`

Since we are calculating the force when the car stops this means that the final velocity is zero:

`0=v_0+at`

Now we solve for the acceleration "a", first by subtracting the initial velocity from both sides:

`-v_0=at`

Now we divide by the time "t":

`-(v_0)/(t)=a`

Now we substitute the values:

`-(28(m)/(s))/(5.8s)=a`

Solving the operations we get:

`-4.83(m)/(s^2)^{}=a`

Now, we use Newton's second law to determine the force:

`F=ma`

Where:

`\begin{gathered} F=\text{ force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}`

Substituting the value we get:

Solving the operations we get:

`F=-6148.59N`

Therefore, the required force is -6148.59 Newtons. The negative sign means that the force is acting in the opposite direction of the movement of the car.