The acceleration experienced by an object will be proportional to the applied force and inversely proportional to its mass. If there are multiple forces, they can be added as vectors and it is the net force that matters.
Key Equations
Newton's Second Law describes his famous equation for the motion of an object
- The change of motion is proportional to the motive force impressed; and is made in the direction of the right (straight) line in which that force is impressed.
The "motion" Newton mentions in the Second Law is, in his language, the product of the mass and velocity of an object --- we call this quantity momentum --- so the Second Law is actually the famous equation: \begin{align*} \vec{F} = \frac{\Delta(m\vec{v})}{\Delta t} = \frac{m\Delta \vec{v}}{\Delta t} =m\vec{a} && \text{[1]} \end{align*}
\begin{align*}\text{Force Sums} \begin{cases} \vec{F_{\text{net}}} = \sum_{i} F_i = m \vec{a} & \text{Net force is the vector sum of all the forces}\\ F_{\text{net,}x} = \sum_{i} {F_{ix}} = m{a}_{x} & \text{Horizontal components add also}\\ F_{\text{net,}y} = \sum_{i} {F_{iy}} = m{a}_{y} & \text{As do vertical ones} \end{cases}\end{align*}
Guidance |
To calculate the net force on an object, you need to calculate all the individual forces acting on the object and then add them as vectors. This requires some mathematical skill. |
Example 1
A 175-g bluebird slams into a window with a force of 19.0 N. What is the bird’s acceleration?
Question: \begin{align*}a = ? [m/s^2]\end{align*}
Given: \begin{align*}m = 175\ grams = 0.175\ kg\end{align*}
\begin{align*}{\;}\qquad \quad F = 19.0 \ N\end{align*}
Equation: \begin{align*}a = \frac{F_{net}}{m}\end{align*}
Plug n’ Chug: \begin{align*}a = \frac{F_{net}}{m} = \frac{19.0 \ N}{0.175 \ kg} = \frac{19.0 \frac{kg \cdot m}{s^2}}{0.175\ kg} = 109 \frac{m}{s^2}\end{align*}
Answer: \begin{align*}\boxed{\mathbf{109 \ m/s^2}}\end{align*}
Example 2
Calculate the acceleration of a rocket that has 500N of thrust force and a mass of 10kg.
Question: \begin{align*}a = ? [m/s^2]\end{align*}
Given: \begin{align*}m = 10\ kg\end{align*}
\begin{align*}{\;} \qquad \quad F_{\text{thrust}} = 500\ N\end{align*}
\begin{align*}{\;} \qquad \quad g = 10.0\ m/s^2\end{align*}
Equations: \begin{align*}\sum F_{\text{individual forces}} = ma\end{align*}
or, in this case, \begin{align*}\sum F_{y-\text{direction forces}} = ma_y\end{align*}
Plug n’ Chug: Use FBD to “fill in” Newton’s second law equation:
\begin{align*}\sum F_{y-\text{direction forces}} &= ma_y \\ F - Mg & = Ma \\ 500N - 10\ kg(10\ m/s^2) & = 10kg (a) \\ a & = 40\ m/s^2\end{align*}
Simulation
Forces in 1 Dimension (PhET Simulation)
Time for Practice
- During a rocket launch, the rocket’s acceleration increases greatly over time. Explain, using Newton’s Second Law. (Hint: most of the mass of a rocket on the launch pad is fuel).
- When pulling a paper towel from a paper towel roll, why is a quick jerk more effective than a slow pull?
- You pull a wagon with a force of 20 N. The wagon has a mas of 10 kg. What is the wagon's acceleration?
- The man is hanging from a rope wrapped around a pulley and attached to both of his shoulders. The pulley is fixed to the wall. The rope is designed to hold 500 N of weight; at higher tension, it will break. Let’s say he has a mass of 80 kg. Draw a free body diagram and explain (using Newton’s Laws) whether or not the rope will break.
- Now the man ties one end of the rope to the ground and is held up by the other. Does the rope break in this situation? What precisely is the difference between this problem and the one before?
- A physics student weighing 500 N stands on a scale in an elevator and records the scale reading over time. The data are shown in the graph below. At time \begin{align*}t = 0\end{align*}, the elevator is at rest on the ground floor.