Chapter 5 of 8

All 8 chapters
  1. 1Scalars and vectors, readunderstood
  2. 2Velocity and acceleration, readunderstood
  3. 3Newton's first law, readunderstood
  4. 4Newton's second law, readunderstood
  5. 5Weight and falling, readunderstood
  6. 6Newton's third law, readunderstood
  7. 7Momentum, readunderstood
  8. 8Stopping distances, readunderstood

S4.8.5 Chapter 5

Weight and falling

A skydiver jumps from a plane. Why does she stop speeding up long before she lands?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Calculate weight from mass and gravitational field strength
  • Explain terminal velocity using the forces on a falling object
  • Time paper cones falling to show terminal velocity

Weight is a force. It is the pull of gravity on an object, so it is a vector that points down. We measure it in newtons. Mass is the amount of matter in an object. We measure it in kilograms.

Each kilogram feels a pull from gravity. The pull on each kilogram is the gravitational field strength, g. Its unit is the newton per kilogram, N/kg.

At the surface of the Earth, g is about 9.8 N/kg. It is a little more near the poles and a little less near the equator, from about 9.78 to 9.83 N/kg. Some books and quick sums use 10 N/kg. This book uses 9.8 N/kg, so answers with 10 will differ a little.

The weight of an object is its mass times g: W = m × g. W is in newtons, m in kilograms and g in N/kg.

See it three ways

As a picture

A block labelled 2 kg with one arrow down labelled weight 19.6 N. A bar marked bar = 10 N shows the scale.2 kgweight 19.6 Nbar = 10 N

In words

Weight is the pull of gravity on an object. It is a force, so it is measured in newtons. Mass is not a force.

In symbols

W = m × g = 2 kg × 9.8 N/kg = 19.6 N

A table with the columns Place and g (N/kg). Earth: 9.8. Moon: 1.6. Mars: 3.7. Jupiter: 25.Placeg (N/kg)Earth9.8Moon1.6Mars3.7Jupiter25
Gravitational field strength at the surface. For Jupiter the value is at its cloud tops, about 25 N/kg.

Worked example

A bag has a mass of 5 kg. Find its weight on Earth, using g = 9.8 N/kg.

  1. Use W = m × g, with m = 5 kg and g = 9.8 N/kg.
  2. W = 5 × 9.8 = 49 N.
  3. The weight is 49 N, acting downwards.

Answer: 49 N

A block labelled 5 kg hanging from a rope. An arrow down is labelled weight 49 N. An arrow up of the same length is labelled rope 49 N. A bar marked bar = 20 N shows the scale.5 kgweight 49 Nrope 49 Nbar = 20 N
The bag hangs still, so the rope pulls up with a force equal to the weight.

Mass stays the same wherever you go. Weight changes with g. An astronaut who goes to the Moon has the same mass and a smaller weight, because g on the Moon is smaller.

Worked example

An astronaut has a mass of 80 kg. Find her weight on Earth and on the Moon, where g is 1.6 N/kg.

  1. Her mass is 80 kg in both places. Only g changes.
  2. On Earth: W = 80 × 9.8 = 784 N.
  3. On the Moon: W = 80 × 1.6 = 128 N.

Answer: Earth 784 N, Moon 128 N

A bar chart called weight (N) for a mass of 80 kg. The bar for Earth reaches 784. The bar for Moon reaches 128. The scale goes from 0 to 800 in steps of 200.0200400600800EarthMoonweight (N)
The same 80 kg of astronaut. The weight on the Moon is much smaller.

Now let an object fall with nothing else acting but its weight. By the second law, a = W ÷ m = (m × g) ÷ m = g. So every object falls with the same acceleration, g = 9.8 m/s². The unit N/kg is the same as m/s².

In 1971 an astronaut on the Moon, where there is no air, dropped a hammer and a feather together. They reached the ground at the same time. On Earth a feather falls slowly because of the air.

Air pushes up on a falling object. We call this force drag, or air resistance. Drag grows as the speed grows. It also depends on the shape and the area of the object. Now follow a skydiver of mass 80 kg, whose weight is 784 N.

Three stages of a falling skydiver, shown as a ball with arrows. Just jumped: weight 784 N, drag 0 N, resultant 784 N down. Speeding up: weight 784 N, drag 392 N, resultant 392 N down. Steady speed: weight 784 N, drag 784 N, resultant 0 N. The weight arrow is the same in each. The drag arrow grows.just jumpedweight 784 Ndrag 0 Nresultant 784 N downspeeding upweight 784 Ndrag 392 Nresultant 392 N downsteady speedweight 784 Ndrag 784 Nresultant 0 N
The weight never changes. The drag grows with speed until it equals the weight.

At the first moment she is hardly falling yet, so there is no upward drag. The resultant is her whole weight, and her acceleration is g, 9.8 m/s².

As she speeds up, the drag grows. The resultant force down gets smaller, so her acceleration gets smaller. She still speeds up, but more slowly.

At last the drag equals the weight. The resultant is 0 N, and her acceleration is 0. She falls at a constant velocity. This is the terminal velocity. A skydiver falling flat and spread out has a terminal velocity of about 55 m/s. It differs from person to person.

Her weight has not changed, and she has not stopped. Gravity still pulls her down. She keeps falling at a constant velocity, because the forces are balanced.

A graph with the axes velocity (m/s) and time (s). Time runs from 0 to 40 s and velocity from 0 to 60 m/s. A curve rises steeply from 0, flattens near 55 m/s by 15 s, then drops after 20 s towards a low steady value of about 5 m/s.5101520253035401020304050600time (s)velocity (m/s)
A model of a jump, not measured data. The gradient shrinks to zero at the first terminal velocity. The parachute opens at 20 s.

When the parachute opens, the area facing the air is much bigger. The drag jumps above the weight. The resultant force now points up, so she slows down.

Two stages of the skydiver after the parachute opens, shown as a ball with arrows. Parachute opens: weight 784 N, drag 1 568 N, resultant 784 N up. Slow, steady speed: weight 784 N, drag 784 N, resultant 0 N.parachute opensweight 784 Ndrag 1 568 Nresultant 784 N upslow, steadyweight 784 Ndrag 784 Nresultant 0 N
As she slows, the drag falls, until it equals the weight again. This is a new, smaller terminal velocity.

As she slows, the drag falls. She stops slowing when the drag again equals the weight. Then she falls at a new, smaller terminal velocity, about 5 m/s with an open parachute. A smooth landing is possible at that speed.

Worked example

A skydiver of mass 80 kg falls at her terminal velocity. Her parachute then opens and the drag jumps to 1 568 N. Find the drag before it opens, and her acceleration just after.

  1. At terminal velocity the resultant force is zero, so the drag equals the weight.
  2. The weight is 80 kg × 9.8 N/kg = 784 N, so the drag was 784 N.
  3. Just after the parachute opens, the resultant is 1 568 N − 784 N = 784 N, upwards.
  4. Her acceleration is 784 N ÷ 80 kg = 9.8 m/s², upwards.

Answer: Before: 784 N. After: 9.8 m/s² upwards

Try it

Time a paper cone as it falls

You need: a sheet of paper, torn into a square, tape, or fold the ends in, a string stuck up a wall, marked at 1 m, 1.5 m and 2 m above the floor, a clock or phone that shows tenths of a second, a friend to time.

Be safe: Stand on the floor and hold the cone up high. Never drop things from a window, a roof or a balcony, and do not stand on a chair.

  1. Roll the paper square into a cone with a point. Tape the edge or tuck it in.
  2. Hold the cone point down, level with the 1 m mark on the string. Let go. Your friend times it from the moment you let go until it lands.
  3. Drop it five times and take the middle time. Then do the same from 1.5 m and from 2 m.
  4. Write down height and time for each. Plot time across and height up.

What you will see: Each extra half metre takes about the same extra time. The cone fell its first stretch while speeding up. Then the drag grew to equal its weight, and it fell at a constant velocity. The gradient of the line is the terminal velocity.

If you cannot try it, this is what happens.
A graph with the axes height fallen (m) and time (s). Time runs from 0 to 1.5 s and height from 0 to 2 m. Three dots lie on a rising line: 1 m at 0.8 s, 1.5 m at 1.1 s and 2 m at 1.4 s.00.511.500.511.52time (s)height fallen (m)

In the example, 0.5 m more takes 0.3 s more. The terminal velocity is about 0.5 ÷ 0.3, which is 1.7 m/s. Your results will differ, but the later part of the line should be straight.

A wrong idea many people have

Mass and weight are the same thing.

Why it feels right: People say "my weight is 60 kilograms", and a kitchen scale marks its dial in kilograms.

The test that breaks it: Take a 60 kg person to the Moon. Her mass is still 60 kg, since she is made of the same matter. Her weight falls from about 588 N to 96 N, because g is smaller there.

What is true: Mass, in kilograms, is how much matter there is, and it does not change with place. Weight, in newtons, is the pull of gravity and it does. Weight = mass × g.

A wrong idea many people have

Heavier objects always fall faster.

Why it feels right: A stone falls faster than a leaf, and a leaf is lighter.

The test that breaks it: Where there is no air, a hammer and a feather fall together. Screw a sheet of paper into a tight ball and drop it with a flat sheet. The ball falls faster, with the same weight.

What is true: With no drag every object falls with the same acceleration, g. With drag, the shape and area matter. A flat sheet has more drag than a ball of the same weight.

A wrong idea many people have

At terminal velocity the skydiver has stopped, or her weight has gone.

Why it feels right: She stops speeding up, and "terminal" sounds like the end.

The test that breaks it: Look at the forces. The weight is still 784 N down. The drag is 784 N up. The resultant is zero, so her velocity is constant, and it is still large.

What is true: At terminal velocity she keeps falling at a constant velocity. The weight is the same, and the drag has grown to equal it.

Key words

weight
the pull of gravity on an object, in newtons. W = m × g.
gravitational field strength
the force of gravity on each kilogram, g, in N/kg. It is 9.8 N/kg on Earth.
drag
the force of air or water pushing against a moving object.
terminal velocity
the constant velocity of a falling object when the drag equals its weight.

Back to your guess

The answer is B: The air pushes up on her harder and harder, until it equals her weight.

The air pushes up harder as she speeds up. When that push equals her weight, the resultant force is zero. She then falls at a constant velocity, called the terminal velocity.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A bag has a mass of 6 kg. What is its weight on Earth, taking g = 9.8 N/kg?

    Tap your answer.

    1. Not this one. That is the mass in kilograms with a newton sign. Multiply by g to get the weight. Try another.

    2. Not this one. You divided the mass by g. The weight is the mass times g. Try another.

    3. Not this one. You added the mass and g. The weight is a product, W = m × g. Try another.

    4. Yes, that is right. W = m × g = 6 kg × 9.8 N/kg = 58.8 N.

    See the answer

    58.8 N

    W = m × g = 6 kg × 9.8 N/kg = 58.8 N.

    • If you chose A: That is the mass in kilograms with a newton sign. Multiply by g to get the weight.
    • If you chose B: You divided the mass by g. The weight is the mass times g.
    • If you chose C: You added the mass and g. The weight is a product, W = m × g.

  2. A girl has a mass of 45 kg on Earth. What is her mass on Mars, where g = 3.7 N/kg?

    Tap your answer.

    1. Yes, that is right. Mass is the amount of matter, so it is the same everywhere. On Mars, W = 45 × 3.7 = 166.5 N, and m = W ÷ g = 166.5 ÷ 3.7 = 45 kg.

    2. Not this one. You scaled the mass with g. Weight changes with g, but mass does not. Go to S3.7.4. Try another.

    3. Not this one. That is her weight on Mars, in newtons, with the wrong unit. Her mass stays the same. Try another.

    See the answer

    45 kg

    Mass is the amount of matter, so it is the same everywhere. On Mars, W = 45 × 3.7 = 166.5 N, and m = W ÷ g = 166.5 ÷ 3.7 = 45 kg.

    • If you chose B: You scaled the mass with g. Weight changes with g, but mass does not. Go to S3.7.4.
    • If you chose C: That is her weight on Mars, in newtons, with the wrong unit. Her mass stays the same.

  3. A skydiver weighs 700 N and falls at a steady terminal velocity. How big is the drag on her?

    Tap your answer.

    1. Not this one. There is plenty of drag. If it were 0 N she would still be speeding up. Try another.

    2. Yes, that is right. A steady velocity means the resultant force is zero. So the drag, up, equals the weight, down: 700 N.

    3. Not this one. With a drag bigger than the weight she would slow down. At a steady velocity the two are equal. Try another.

    See the answer

    700 N

    A steady velocity means the resultant force is zero. So the drag, up, equals the weight, down: 700 N.

    • If you chose A: There is plenty of drag. If it were 0 N she would still be speeding up.
    • If you chose C: With a drag bigger than the weight she would slow down. At a steady velocity the two are equal.

  4. A hammer and a feather are dropped together on the Moon, where there is no air. What happens?

    Tap your answer.

    1. Not this one. The hammer has a bigger weight, but also more mass. The acceleration W ÷ m is g for both. Try another.

    2. Not this one. A lighter object does not fall faster. With no air, both have the same acceleration. Try another.

    3. Yes, that is right. With no air there is no drag. Each object has a = W ÷ m = g, so they fall together.

    See the answer

    They land together, because they have the same acceleration.

    With no air there is no drag. Each object has a = W ÷ m = g, so they fall together.

    • If you chose A: The hammer has a bigger weight, but also more mass. The acceleration W ÷ m is g for both.
    • If you chose B: A lighter object does not fall faster. With no air, both have the same acceleration.

  5. A skydiver opens her parachute and the drag becomes bigger than her weight. What happens next?

    Tap your answer.

    1. Yes, that is right. The drag, up, is bigger than the weight, down. The resultant points up, opposite to her velocity, so she slows down.

    2. Not this one. That needs the drag to equal the weight. Here the drag is bigger. Try another.

    3. Not this one. A resultant that points up cannot speed up a fall. It slows her down. Try another.

    See the answer

    She slows down.

    The drag, up, is bigger than the weight, down. The resultant points up, opposite to her velocity, so she slows down.

    • If you chose B: That needs the drag to equal the weight. Here the drag is bigger.
    • If you chose C: A resultant that points up cannot speed up a fall. It slows her down.

  6. Why does a falling skydiver reach a terminal velocity?

    Tap your answer.

    1. Not this one. Gravity is almost the same over the whole fall. It is the drag that changes. Try another.

    2. Yes, that is right. Drag grows as she speeds up. When it equals her weight the resultant is zero, so her velocity stays constant.

    3. Not this one. Her mass does not change. The forces change, because the drag grows. Try another.

    See the answer

    The drag grows with her speed until it equals her weight, so the resultant force is zero.

    Drag grows as she speeds up. When it equals her weight the resultant is zero, so her velocity stays constant.

    • If you chose A: Gravity is almost the same over the whole fall. It is the drag that changes.
    • If you chose C: Her mass does not change. The forces change, because the drag grows.

  7. A paper cone falls 1 m in 0.9 s, 1.5 m in 1.15 s and 2 m in 1.4 s. What does this show?

    Tap your answer.

    1. Yes, that is right. Each extra half metre takes the same 0.25 s. Equal distances in equal times mean a constant velocity. The drag equals the weight, so the resultant force is zero.

    2. Not this one. If it were speeding up, each extra half metre would take less time than the one before. Try another.

    3. Not this one. The weight is the same all the way down. The drag has grown until it equals the weight. Try another.

    See the answer

    It falls at a constant velocity, its terminal velocity.

    Each extra half metre takes the same 0.25 s. Equal distances in equal times mean a constant velocity. The drag equals the weight, so the resultant force is zero.

    • If you chose B: If it were speeding up, each extra half metre would take less time than the one before.
    • If you chose C: The weight is the same all the way down. The drag has grown until it equals the weight.

  8. From an earlier chapter

    A car of mass 1 200 kg has a resultant force of 3 000 N on it. What is its acceleration?

    Tap your answer.

    1. Not this one. You divided the mass by the force. It is the force divided by the mass. Go to S4.8.4. Try another.

    2. Not this one. You added the force and the mass. The second law is a division, a = F ÷ m. Go to S4.8.4. Try another.

    3. Yes, that is right. a = F ÷ m = 3 000 N ÷ 1 200 kg = 2.5 m/s².

    See the answer

    2.5 m/s²

    a = F ÷ m = 3 000 N ÷ 1 200 kg = 2.5 m/s².

    • If you chose A: You divided the mass by the force. It is the force divided by the mass. Go to S4.8.4.
    • If you chose B: You added the force and the mass. The second law is a division, a = F ÷ m. Go to S4.8.4.

  9. A rock has a mass of 12 kg. What is its weight on the Moon, where g = 1.6 N/kg?

    See the answer

    19.2 N

    W = m × g = 12 kg × 1.6 N/kg = 19.2 N.

  10. A parachutist of mass 55 kg falls at a steady terminal velocity. How big is the drag on her? Use g = 9.8 N/kg.

    See the answer

    539 N

    The weight is 55 kg × 9.8 N/kg = 539 N. At a steady velocity the resultant is zero, so the drag is 539 N.

  11. A paper cone falls at its terminal velocity. It falls an extra 1.5 m in 0.6 s. What is its terminal velocity?

    See the answer

    2.5 m/s

    Speed = distance ÷ time = 1.5 m ÷ 0.6 s = 2.5 m/s.

Keep it

Copy this into your exercise book.

Weight W = m × g, in newtons, with g = 9.8 N/kg on Earth. Mass does not change. A falling object has terminal velocity when drag = weight and the resultant is zero.

Draw: A falling ball with a weight arrow down and a drag arrow up growing until they are equal.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (825 KB). It opens on any phone.