Question

A 1,052 kg car moving at 84.21 km/h comes to a stop in 9.260s. The magnitude of the net force on the can in Newtons is?

Answer 1

First, we have to find the acceleration of the car with the following formula.

$v_f=v_0+at$

Where vf = 0, v0 = 84.21 km/h, and t = 9.260 s. Let's use these values and solve for a.

$\begin{gathered} 0=84.21(\frac{km}{h})+a\cdot9.26\sec \\ -84.21(\frac{km}{h})=a\cdot9.26\sec \\ a=\frac{-84.21(\frac{km}{h})}{9.26\sec } \end{gathered}$

Before we divide, we have to express the speed in meters per second. We know that 1 km equals 1000 meters, and 1 hour equals 3600 seconds.

$\frac{84.21\operatorname{km}}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600\sec }\approx23.39(\frac{m}{s})$

Then, we use the transformed speed to find the acceleration.

$a=\frac{-23.39(\frac{m}{s})}{9.26\sec }\approx-2.53(\frac{m}{s^2})$

Once we have the acceleration, we can find the net force using Newton's Second Law.

$F=ma$

Where m = 1,052 kg and a = - 2.53 m/s2.

$F=1052\operatorname{kg}\cdot(-2.53(\frac{m}{s^2}))=-2661.56N$

Therefore, the net force is 2661.56 Newtons.

The negative sign indicates that the force applied is developing a negative acceleration on the car.