Question

A 873-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time?_____ kN eastward

Answer 1

Given:

The mass of the car is: m = 873 kg

The initial speed of the car is: vi = 0 m/s (as initially, the car is at rest)

The final speed of the car is: vf = 35 m/s

The time over which the speed of car increases is: t = 5 s

To find:

The force exerted on the car.

Step-by-step explanation:

The acceleration of the car can be calculated by using the following equation.

`a=(v_f-v_i)/(t)`

Here, a is the acceleration of the car.

Substituting the values in the above equation, we get:

`\begin{gathered} a=\frac{35\text{ m/s}-0\text{ m/s}}{5\text{ s}} \\ \\ a=\frac{35\text{ m/s}}{5\text{ s}} \\ \\ a=7\text{ m/s}^2 \end{gathered}`

The acceleration of the car is 7 m/s² in an eastward direction.

Now, using Newton's second law of motion, the force exerted on the car can be calculated as:

`F=ma`

Substituting the values in the above equation, we get:

`\begin{gathered} F=873\text{ kg}*7\text{ m/s}^2 \\ \\ F=6111\text{ N} \\ \\ F=6111*\frac{\text{ kN}}{10^3} \\ \\ F=6.111\text{ kN} \end{gathered}`

The force exerted on the car is 6.111 kN. The direction of force is the same as the direction of the car. Thus, the direction of the force is in an eastward direction.

Final answer:

The force exerted on the car is 6.111 Kilo-Newtons (kN) eastward.