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Now let’s work an example of Newton’s Second Law with numbers.

Two forces act on an object with a mass of 2.5 kg.  Suppose that $$\color{black}{\vec{F}_1 = -12 \; \mathrm{N}\; \hat{x}}$$ and $$\color{black}{\vec{F}_2 = 7 \; \mathrm{N} \; \hat{x}}$$.  What is the acceleration of the object?

## Setting up the Problem First, we draw the free body diagram for the situation.  We make a mental note that the final acceleration should be negative.  If our answer isn’t negative, it is likely that we have made a mistake.

In the last lesson, we went from the working form of Newton’s Second Law through the vector equation to an algebra equation.  These steps looked like…

$$\color{black}{m \vec{a} = \Sigma \vec{F}}$$

$$\color{black}{m \vec{a} = \vec{F}_1 + \vec{F}_2}$$

## Moving to the Algebra Equation and Solving

Now we change from the vector equation to the algebra equation.  We put in signs for the forces, but not the acceleration.  We are looking for the acceleration, so we solve for it, and allow the algebra and equation to determine the sign.

$$\color{black}{m a = {}^-F_1 + {}^+F_2}$$

$$\color{black}{a = \cfrac{{}^-F_1 + {}^+F_2}{m}}$$

## Putting in the Numbers

Next, we put in the values of the magnitudes of the forces and the mass.  The signs have been taken care of by the equation.  If we put them in again, we will have double corrected.  This can lead to incorrect answers.

$$\color{black}{a = \cfrac{{}^-12 \;\mathrm{N}\; + {}^+7 \;\mathrm{N}}{2.5 \;\mathrm{kg}}}$$

You can either enter this directly into your calculator, or work out a few more steps by hand.  I recommend that you do it by hand so that you can check units along the way.

$$\color{black}{a =\cfrac{{}^-5 \;\mathrm{\frac{kg m}{s^2}}}{2.5\; \mathrm{kg}}}$$

We broke the newton into its SI bits.  It is now obvious that the kg cancel out from the numerator and denominator.  This leaves us with…

$$\color{black}{a = -2.0\; \mathrm{\cfrac{m}{s^2}}}$$

The fact that the acceleration is negative means that the object will accelerate to the left. This matches with our free body diagram, in which it is obvious that the net force (and therefore acceleration) must point to the left.