Newton's second law
The first law tells you what happens when the net force is zero: nothing changes. The second law tells you what happens when it is not zero, and by how much. It is the most used equation in mechanics.
By the end of this lesson you can
- describe how acceleration depends on net force and on mass;
- use \(\sum \vec F = m\vec a\) to find a force, a mass or an acceleration;
- predict the direction of an acceleration from the net force;
- combine the second law with the equations of motion.
Force, mass and acceleration
Think first
(a) You push a stalled car alone, then three friends join you. What happens to how quickly it speeds up? (b) You push a small car and then a loaded van with the same force. Which speeds up faster?
Compare your answer
(a) With four people the net force is about four times larger, and the car speeds up about four times as quickly. (b) The small car. With the same force, more mass means less acceleration.
These two everyday observations are the whole content of the second law:
- More net force, more acceleration, in the same proportion: double one and you double the other.
- More mass, less acceleration, in inverse proportion: double the mass and the acceleration halves.
(Friction is ignored here. In real life, pushing a car only speeds it up once your push is bigger than the friction holding it back. That is exactly why the law talks about the net force.)
Change the force, change the mass
InteractiveScroll sideways to see the whole demo, or turn your phone sideways.
Try this
- Set the mass to 2 kg and the force to 4 N, and press Go. Note the acceleration.
- Double the force to 8 N. What happens to the acceleration?
- Keep 8 N but double the mass to 4 kg. Explain the result.
Newton's second law
Newton's second law
$$\sum \vec F = m\vec a$$Multiply an object's mass by its acceleration and you get the net force on it. Force and acceleration always point the same way.
The Greek letter \(\sum\) stands for "add up everything", so \(\sum \vec F\) is the net force. Because force and acceleration are vectors, the law holds along each axis separately:
$$\sum F_x = ma_x \qquad \sum F_y = ma_y$$Rearranged, \(a = \dfrac{\sum F}{m}\) shows both proportions at once: force on top, mass underneath. When the net force is zero, the acceleration is zero, which is just the first law again.
Units check
With mass in kilograms and acceleration in m/s², force comes out in newtons, since \(1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2\). Convert grams to kilograms before you substitute.
Worked example 1
A 1200 kg car has a net force of 3000 N acting on it, toward the east. What is its acceleration?
- List what you know. \(m = 1200\ \text{kg}\), \(\sum F = 3000\ \text{N}\) east. Unknown: \(a\).
- Rearrange the second law. \(a = \dfrac{\sum F}{m}\)
- Substitute. $$a = \frac{3000\ \text{N}}{1200\ \text{kg}} = 2.5\ \text{m/s}^2$$
- Give the direction. The acceleration points the same way as the net force.
\(a = 2.5\ \text{m/s}^2\) east
Worked example 2
A 55 kg skier starts from rest and slides 36 m down a slope in 6.0 s with constant acceleration. What is the net force on the skier along the slope?
- Plan. You cannot use \(\sum F = ma\) yet, because you do not know \(a\). Find \(a\) from the motion first, then the force.
- Find the acceleration. Starting from rest, \(\Delta x = \tfrac12 a(\Delta t)^2\), so $$a = \frac{2\Delta x}{(\Delta t)^2} = \frac{2(36\ \text{m})}{(6.0\ \text{s})^2} = 2.0\ \text{m/s}^2$$
- Apply the second law. $$\sum F = ma = (55\ \text{kg})(2.0\ \text{m/s}^2) = 110\ \text{N}$$
\(\sum F = 110\ \text{N}\) down the slope
Worked example 3
Two players hit a 2.0 kg puck on frictionless ice at the same moment: one with 6.0 N toward the east, the other with 8.0 N toward the north. Find the puck's acceleration.
- Find the net force first. The forces are perpendicular: \(\sum F = \sqrt{(6.0\ \text{N})^2 + (8.0\ \text{N})^2} = 10\ \text{N}\).
- Find its direction. \(\theta = \tan^{-1}\left(\dfrac{8.0}{6.0}\right) = 53^\circ\) north of east.
- Apply the second law. \(a = \dfrac{10\ \text{N}}{2.0\ \text{kg}} = 5.0\ \text{m/s}^2\), in the direction of the net force.
\(a = 5.0\ \text{m/s}^2\) at 53° north of east
Thinking with the second law
Puzzle 1
A 7 kg bowling ball weighs about 70 times as much as a 100 g apple. Yet if you drop both from a balcony, with air resistance negligible, they land together. How can a much bigger force give the same acceleration?
Show explanation
The bowling ball's weight is 70 times larger, but so is its mass. In \(a = F/m\) the two factors of 70 cancel. Gravity's pull always grows in step with the inertia it has to overcome, so everything falls with the same acceleration when air resistance can be ignored.
Puzzle 2
A rocket's engine gives a constant thrust as it burns through its fuel. Ignoring gravity and air, does the rocket's acceleration stay the same, rise or fall during the burn?
Show explanation
It rises. The force is constant, but the rocket gets lighter as fuel is used up, so \(a = F/m\) keeps growing. A rocket is moving fastest and accelerating hardest just before its fuel runs out.
Check yourself
Question 1
A 2.5 kg model rocket has a net upward force of 6.0 N on it. What is its acceleration?
Show solution
\(a = \dfrac{6.0\ \text{N}}{2.5\ \text{kg}} = 2.4\ \text{m/s}^2\) upward.
Question 2
A net force of 12 N gives a cart an acceleration of 1.5 m/s². What is the cart's mass?
Show solution
\(m = \dfrac{\sum F}{a} = \dfrac{12\ \text{N}}{1.5\ \text{m/s}^2} = 8.0\ \text{kg}\)
Question 3
During launch, astronauts can feel an acceleration of about 40 m/s². What net force acts on a 70 kg astronaut?
Show solution
\(\sum F = ma = (70\ \text{kg})(40\ \text{m/s}^2) = 2800\ \text{N}\), about four times the astronaut's weight.
Question 4
A net force \(F\) gives a mass \(m\) an acceleration \(a\). What acceleration does a net force \(2F\) give a mass \(4m\)?
- \(8a\)
- \(2a\)
- \(a\)
- \(a/2\)
Show solution
\(a' = \dfrac{2F}{4m} = \dfrac{1}{2}\cdot\dfrac{F}{m} = \dfrac{a}{2}\)
Question 5
Some net force speeds a 6.0 kg box up at 1.5 m/s². What acceleration would that same net force give a 3.0 kg box?
Show solution
First the force: \(\sum F = (6.0\ \text{kg})(1.5\ \text{m/s}^2) = 9.0\ \text{N}\). Then \(a = \dfrac{9.0\ \text{N}}{3.0\ \text{kg}} = 3.0\ \text{m/s}^2\). Half the mass, twice the acceleration.
Question 6
A 1.5 kg ball, dropped from rest, falls 4.9 m in 1.0 s. Find the net force on the ball.
Show solution
\(a = \dfrac{2\Delta x}{(\Delta t)^2} = \dfrac{2(4.9\ \text{m})}{(1.0\ \text{s})^2} = 9.8\ \text{m/s}^2\), so \(\sum F = (1.5\ \text{kg})(9.8\ \text{m/s}^2) = 14.7\ \text{N} \approx 15\ \text{N}\).
Summary
- Acceleration grows with net force and shrinks with mass.
- Newton's second law: \(\sum \vec F = m\vec a\). It also holds along each axis: \(\sum F_x = ma_x\), \(\sum F_y = ma_y\).
- An object accelerates in whatever direction the net force points.
- With several forces, add them up before using the law. If the acceleration is not given, get it from the motion.
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