2.5 Newton's Second Law

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. An elevator ride on a bathroom scale

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You step into an elevator on the ground floor carrying a bathroom scale. You put the scale on the floor and stand on it. It reads 588 N (that is 60 kg). The doors close and the elevator starts going up.

For a moment the number jumps to about 660 N. You feel heavier. Then, while the elevator glides up, the number goes back to 588 N. Near the top floor it drops to about 516 N and you feel light. Your mass never changed.

The question this topic answers: how do the forces on an object decide its acceleration, and why did the scale change its reading?

2. An empty cart and a full cart

3. Tug of war on a wagon

Quick check. Sam pushes a cart with the same force as before, but now the cart has three times the mass. What happens to its acceleration?

4. Braking on a bike

5. Pushing a box at a steady speed

6. Two boxes, one system

Quick check. Leo's bike still rolls forward while he brakes. Which way does the net force on Leo and his bike point?

7. Check yourself

Think of your answer first, then tap to see it.

a) A car moves at a steady 25 m/s on a straight highway. What is the net force on it?

Show answer

Zero. Its velocity is not changing, so a = 0 and Fnet = m a = 0.

b) The same net force acts on a cart, then on a cart with twice the mass. How do their accelerations compare?

Show answer

The heavier cart gets half the acceleration, because a = Fnet ÷ m.

c) A 1200 kg car has a net force of 3600 N forward. What is its acceleration?

Show answer

a = 3600 N ÷ 1200 kg = 3 m/s² forward.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. A hockey puck slides across smooth ice in a straight line at a steady speed. What is the net force on it?

2. A net force of 12 N acts on a 3.0 kg box. What is its acceleration?

3. You stand on a scale in an elevator that moves down at a steady speed. Compared with your weight, the scale reads:

The idea in plain words

Add up all the forces on one object (the system). The total is the net force. The net force makes the object accelerate. A bigger net force gives a bigger acceleration. A bigger mass gives a smaller acceleration for the same net force.

a = Fnet / m    or    Fnet = m a

Trap: putting a weight in newtons where the mass goes (a 588 N person used as m = 588). Instead: m is in kilograms; if you are given a weight, divide by g first: m = 588 N / 9.8 m/s² = 60 kg.

Three things to remember:

Trap: drawing "ma" as an extra arrow on the free body diagram. Instead: draw only real forces from real objects; ma is what they add up to, on the other side of Fnet = ma.

Apparent weight. A scale does not measure your weight. It measures how hard it pushes up on you (the normal force FN). With up positive, the forces on you are FN up and mg down, so FN − mg = ma, which gives FN = m(g + a). If the elevator accelerates up (a > 0), the scale reads more than mg. If it accelerates down (a < 0), it reads less. At constant velocity, a = 0 and the scale reads exactly mg, even while moving fast.

Trap: thinking a scale always reads your weight mg. Instead: it reads the normal force, FN = m(g + a); that equals mg only when a = 0.

Worked numbers: the scale reading in the story

Read the steps as text

You have mass 60 kg. Up is positive. g = 9.8 m/s².

  1. Choose the system: you. Draw your free body diagram: FN (scale, up) and Fg = mg (Earth, down). Why: the scale reading is FN, so we need a diagram that has it on it.
  2. Your weight: Fg = mg = 60 kg × 9.8 m/s² = 588 N. Why: this is the reading at rest, when a = 0.
  3. Write Newton's second law with up positive: FN − mg = ma. Why: FN points up (+), gravity points down (−).
  4. Speeding up going up: a = +1.2 m/s². FN = m(g + a) = 60 × (9.8 + 1.2) = 60 × 11.0 = 660 N. Why: the floor must push harder than gravity pulls, so the net force points up.
  5. Gliding up at steady speed: a = 0. FN = 60 × 9.8 = 588 N. Why: constant velocity means zero net force, so FN balances gravity.
  6. Slowing down near the top: the velocity is up but it is shrinking, so a = −1.2 m/s². FN = 60 × (9.8 − 1.2) = 60 × 8.6 = 516 N. Why: slowing an upward motion needs a downward net force, so the floor pushes less than gravity.
Trap: using a positive a while the elevator slows down near the top, because it is still moving up. Instead: slowing down means a points opposite to v; moving up and slowing gives a = −1.2 m/s², so FN = 516 N.

Lab 1: a fan cart

A small fan on a cart pushes the cart along a level, frictionless track (positive is to the right). Change the fan force, the mass and the starting velocity. Predict the velocity graph first, then press play.

Gravity and the track's normal force cancel. The fan force is the net force, so a = Ffan / m. Double the fan force and the acceleration doubles. Double the mass and it halves. Try a negative fan force with a positive starting velocity: the cart slows, stops and comes back.

Lab 2: the elevator and the scale

A person stands on a scale in an elevator. The elevator speeds up, cruises at steady speed, then slows down and stops. Up is positive. Set the trip direction to +1 for up or −1 for down. Predict how the scale reading changes.

The scale reading is the normal force: FN = m(g + a). It depends on the acceleration, not on the velocity. During the cruise the elevator may be moving fast, but the reading is exactly mg.

Lab 3: the Atwood machine

Two masses hang from a light string over a light, frictionless pulley. Mass 1 is on the left, mass 2 on the right. Both start at rest at the same height. The graphs follow mass 2, with up positive.

Treat both masses and the string as one system: the outside force that drives it is the difference in weights, (m2 − m1)g, and the mass being accelerated is m1 + m2. So a = (m2 − m1)g / (m1 + m2). The tension is T = 2m1m2g / (m1 + m2): always between the two weights when the masses differ.

Examples

Example 1: a car speeding up basic

The net force on a 1500 kg car is 3000 N forward. What is its acceleration?

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  1. a = Fnet / m = 3000 N / 1500 kg = 2.0 m/s² forward. Why: 1 N = 1 kg·m/s², so N / kg gives m/s². The acceleration points the same way as the net force.

Example 2: the elevator scale medium

A 60 kg student stands on a scale in an elevator. Find the scale reading when the elevator (a) speeds up going up at 2.0 m/s² and (b) speeds up going down at 2.0 m/s². (c) A scale marked in kilograms divides FN by 9.8. What does it show in (a)?

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  1. Up positive. Forces on the student: FN up, mg = 588 N down. FN − mg = ma.
  2. (a) a = +2.0 m/s²: FN = 60(9.8 + 2.0) = 708 N. Why: speeding up upward means a points up.
  3. (b) Speeding up downward means a points down: a = −2.0 m/s². FN = 60(9.8 − 2.0) = 468 N.
  4. (c) 708 N / 9.8 m/s² = 72 kg. The scale "thinks" the student is 72 kg, but the mass is still 60 kg.

Example 3: two blocks and a string medium

On a frictionless floor, a 2.0 kg block is pulled to the right by a 20 N force. A string connects it to a 3.0 kg block behind it. Find the acceleration and the tension in the string.

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  1. System: both blocks. The only horizontal external force is 20 N. a = 20 N / 5.0 kg = 4.0 m/s² right. Why: the tension is internal to this system, so it cancels.
  2. System: the 3.0 kg block alone. The only horizontal force is the tension. T = ma = 3.0 × 4.0 = 12 N. Why: the back block is pulled only by the string.
  3. Check with the front block: 20 N − 12 N = 8 N, and 2.0 kg × 4.0 m/s² = 8 N. It matches.

Example 4: the Atwood machine AP

A 2.0 kg mass and a 3.0 kg mass hang over a light frictionless pulley and are released from rest. The 3.0 kg mass starts 1.0 m above the floor. Find (a) the acceleration, (b) the tension, (c) how long it takes the 3.0 kg mass to reach the floor and how fast it is going then.

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  1. (a) Positive is "3.0 kg mass moving down, 2.0 kg mass moving up". System: both masses and the string. Driving force: 3.0(9.8) − 2.0(9.8) = 9.8 N. Total mass 5.0 kg. a = 9.8 / 5.0 = 1.96 m/s². Why: the tension pulls both masses toward the pulley; along the string it is internal and cancels.
  2. (b) The 2.0 kg mass alone, up positive: T − 2.0(9.8) = 2.0(1.96), so T = 19.6 + 3.92 = 23.5 N. Check with the formula: 2(2.0)(3.0)(9.8)/5.0 = 23.5 N. It is between 19.6 N and 29.4 N, as it must be.
  3. (c) Starting from rest: 1.0 = ½(1.96)t², t² = 1.02, t = 1.0 s. Speed v = at = 1.96 × 1.01 = 2.0 m/s. Why: constant acceleration, so the Unit 1 equations apply.
Trap: setting the tension equal to one hanging weight (T = 19.6 N or 29.4 N). Instead: the masses accelerate, so T is between the two weights; write Fnet = ma for one mass: T = 23.5 N.

Example 5: a cart pulled by a hanging mass AP

A 4.0 kg cart on a level frictionless table is tied to a string that runs over a pulley at the table's edge to a 1.0 kg hanging mass. Find the acceleration and the tension. Then explain, without numbers, why the tension is less than the hanging mass's weight.

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  1. System: cart + hanging mass + string. The only external force along the string's path that is not balanced is the hanging weight, 1.0 × 9.8 = 9.8 N. Total mass 5.0 kg. a = 9.8 / 5.0 = 1.96 m/s². Why: the cart's weight is balanced by the table's normal force.
  2. Cart alone: the only horizontal force is the tension. T = 4.0 × 1.96 = 7.84 N ≈ 7.8 N.
  3. Check the hanging mass, down positive: 9.8 − 7.84 = 1.96 N, and 1.0 × 1.96 = 1.96 N. It matches.
  4. Why T < mg: the hanging mass accelerates downward, so the net force on it must point down. That only happens if the upward tension is smaller than its downward weight. If T equalled mg, the hanging mass could not speed up.

Practice (AP style)

1. A 3.0 kg box has a net force of 12 N to the left on it. What is its acceleration?

Show answer

a = Fnet/m = 12 / 3.0 = 4.0 m/s², in the direction of the net force (left).

2. The same net force acts on cart X and on cart Y. Cart Y has twice the mass of cart X. How does Y's acceleration compare with X's?

Show answer

a = F/m. Same F and double m gives half the acceleration.

3. A 50 kg person stands on a scale in an elevator that moves downward at a constant 3.0 m/s. What does the scale read?

Show answer

Constant velocity means a = 0, so FN = mg = 50 × 9.8 = 490 N. The speed of 3.0 m/s does not matter.

4. The same 50 kg person rides an elevator that is moving up and slowing down at 1.8 m/s². What does the scale read?

Show answer

Moving up but slowing down means the acceleration points down: a = −1.8 m/s² (up positive). FN = m(g + a) = 50(9.8 − 1.8) = 50 × 8.0 = 400 N.

5. A 1.0 kg mass and a 3.0 kg mass hang on opposite sides of an ideal Atwood machine. What is the size of their acceleration?

Show answer

a = (m2 − m1)g/(m1 + m2) = (2.0)(9.8)/4.0 = 4.9 m/s².

6. In an Atwood machine with unequal masses m1 < m2, released from rest, how does the string tension T compare with the weights?

Show answer

The light mass speeds up upward, so T must be bigger than m1g. The heavy mass speeds up downward, so T must be smaller than m2g. Tension is between the two weights.

7. The velocity-time graph of a 0.50 kg cart is a straight line from 0 m/s at t = 0 to 6.0 m/s at t = 3.0 s. What is the net force on the cart?

Show answer

The slope of v-t is the acceleration: (6.0 − 0)/3.0 = 2.0 m/s². Fnet = ma = 0.50 × 2.0 = 1.0 N.

8. A hockey puck slides to the right. At this moment the net force on it points to the left. What is happening to the puck?

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The acceleration points the same way as the net force: left. The velocity points right. When velocity and acceleration point opposite ways, the object slows down.

9. (Short free response: translating between representations) A 60 kg student stands on a scale in an elevator that starts at rest. The scale reads 660 N for the first 2.0 s, 588 N for the next 5.0 s and 516 N for the last 2.0 s, after which the elevator is at rest again. (a) Find the acceleration in each part, up positive. (b) Is the elevator going up or down? Explain. (c) Find the top speed and the total distance travelled.

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(a) a = (FN − mg)/m with mg = 588 N. Part 1: (660 − 588)/60 = +1.2 m/s². Part 2: 0. Part 3: (516 − 588)/60 = −1.2 m/s².

(b) Up. It started at rest and its first acceleration was upward, so it began moving upward. It cruised, then the downward acceleration slowed it to a stop.

(c) Top speed: v = 1.2 × 2.0 = 2.4 m/s. Distance: speeding up ½(1.2)(2.0)² = 2.4 m; cruising 2.4 × 5.0 = 12 m; slowing down 2.4 m (the mirror image). Total 16.8 m.

10. (Short free response: experimental design) You have a fan cart whose fan force can be set to 1, 2, 3 or 4 equal settings, extra 0.25 kg bars that stack on the cart, a level low-friction track and a motion sensor. Describe an experiment that shows (a) a is proportional to Fnet and (b) a is inversely proportional to m. Say what you keep the same, what you measure and what graph you would make.

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(a) Keep the mass the same. Run the cart at fan settings 1, 2, 3, 4. For each run, use the motion sensor's v-t graph and take the slope as the acceleration. Graph a versus fan setting (proportional to Fnet). A straight line through the origin shows a ∝ Fnet.

(b) Keep the fan setting the same. Add 0, 1, 2, 3 bars and measure the total mass on a balance. Find a for each. Graph a versus 1/m. A straight line through the origin shows a ∝ 1/m. (Graphing a versus m gives a curve, which is harder to judge.)

Good practice: repeat each run several times and average; check the track is level (a cart with the fan off should not speed up); keep the fan's force horizontal.

11. (Short free response: mathematical routines) A 1200 kg car starts from rest on a level road and reaches 24 m/s in 8.0 s with constant acceleration. Air drag and rolling friction together push back with 400 N. (a) Calculate the acceleration. (b) Calculate the forward force from the road on the tires. (c) Calculate the distance covered in the 8.0 s. (d) Calculate the forward force needed to cruise at a steady 24 m/s.

Show answer
  1. (a) a = Δv / Δt = 24 / 8.0 = 3.0 m/s².
  2. (b) Fnet = ma: F − 400 = 1200 × 3.0 = 3600, so F = 4000 N.
  3. (c) Δx = ½ a t² = ½ × 3.0 × 8.0² = 96 m.
  4. (d) Steady speed: a = 0, so F = 400 N (it only cancels the backward forces).

Point guide (4 points):

  • 1 point: a = 3.0 m/s².
  • 1 point: Fnet = F − 400 with F = 4000 N.
  • 1 point: 96 m.
  • 1 point: 400 N with the a = 0 reason.

12. (Short free response: qualitative/quantitative translation) An 800 kg elevator car hangs from a cable with tension T. Up is positive. (a) Derive an expression for T in terms of m, g and a. (b) Without numbers, rank T for these cases, largest first: (1) speeding up going up; (2) moving down at constant speed; (3) slowing down going down; (4) slowing down going up. (c) Calculate T for cases 1 to 4 with |a| = 1.5 m/s², and check your ranking.

Show answer
  1. (a) Forces on the car: T up, mg down. T − mg = ma, so T = m(g + a).
  2. (b) Cases 1 and 3 both have a pointing up (a > 0), case 2 has a = 0, case 4 has a pointing down. Ranking: 1 = 3 > 2 > 4.
  3. (c) Cases 1 and 3: T = 800(9.8 + 1.5) = 9040 N. Case 2: T = 800 × 9.8 = 7840 N. Case 4: T = 800(9.8 − 1.5) = 6640 N. The ranking matches.
  4. Note: case 3 (moving down but slowing) needs as much tension as case 1. The direction of the velocity does not matter; only the acceleration does.

Point guide (4 points):

  • 1 point: T = m(g + a) from a correct free body diagram.
  • 1 point: ranking 1 = 3 > 2 > 4.
  • 1 point: correct sign of a for each case (slowing down going down means a up).
  • 1 point: 9040 N, 7840 N, 6640 N.

Common mistakes

The mistake: drawing "ma" as an extra force on the free body diagram.

Why it is wrong: ma is not a force from any object. It is what the real forces add up to.

How to spot it: every arrow on a free body diagram must name the object that exerts it (Earth, floor, rope). "ma" has no such object, so it does not belong.

The mistake: "A scale always reads my weight."

Why it is wrong: a scale reads the normal force, FN = m(g + a). It equals mg only when a = 0.

How to spot it: ask "is the person accelerating?" If yes, the reading is not mg.

The mistake: setting the tension equal to mg in an accelerating Atwood machine or hanging mass.

Why it is wrong: if T = mg on a hanging mass, its net force is zero and it cannot accelerate.

How to spot it: for a mass speeding up downward, T must come out smaller than its weight; for one speeding up upward, larger.

The mistake: thinking the acceleration (or the net force) points the way the object moves.

Why it is wrong: a points the way the net force points. A car braking moves forward with a backward acceleration.

How to spot it: if the object is slowing down, its acceleration must point opposite its velocity.

The mistake: losing the signs: writing FN + mg = ma, or using a = +2 when the elevator slows while going up.

Why it is wrong: forces and accelerations are vectors; the sign carries the direction.

How to spot it: state "up is positive" in writing before the first equation, and check the answer: an upward acceleration must give FN > mg.

More explanations: free resources.