Question

as stated in the first law, the presence of an unbalanced force will accelerate an object - changing either its speed, its direction, or both its speed and direction. newtons second law of motion pertains to the behavior of objects for which all existing forces are not balanced. the second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. the acceleration is always in the same direction as the net force. mathematically this means: acceleration = \\(\frac{\text{force}_{\text{net}}}{\text{mass}}\\); commonly written as: \\(f_{\text{net}} = ma\\). do not use the value of merely any force in the above equation. it is the net force which is related to acceleration. the net force is the sum of all the forces acting on an object. critical thinking questions – part iii 1. what two variables is acceleration dependent on? what is the relationship between these variables and acceleration? (i.e. if you increase one variable what happens to the acceleration?) 2. if an object is not accelerating what can you determine about the sum of all the forces on the object? 3. if the net force on an object is in a negative direction, what will the direction of the resulting acceleration be? 4. if you double the net force on an object what is the result on the acceleration? 5. if you double the mass of an object while leaving the net force unchanged what is the result on the acceleration? 6. a cadillac escalade has a mass of 2569.6 kg, if it accelerates at 4.65 m/s² what is the net force on the car?

Explanation:

Question 6 Solution:

Step1: Recall Newton's second law formula

We know from Newton's second law that the formula relating net force (\(F_{net}\)), mass (\(m\)) and acceleration (\(a\)) is \(F_{net}=ma\).

Step2: Identify given values

The mass of the Cadillac Escalade, \(m = 2569.6\space kg\) and the acceleration, \(a=4.65\space m/s^{2}\).

Step3: Substitute values into the formula

Substitute \(m = 2569.6\space kg\) and \(a = 4.65\space m/s^{2}\) into the formula \(F_{net}=ma\).
\[

$$\begin{align*} F_{net}&=2569.6\space kg\times4.65\space m/s^{2}\\ &=2569.6\times4.65\space N\\ & = 11948.64\space N \end{align*}$$

\]

Answer:

The net force on the car is \(11948.64\space N\) (or approximately \(1.19\times 10^{4}\space N\) depending on significant figures, but as per calculation it is \(11948.64\space N\)).