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ESO Y4B · Physics and Chemistry
Interaction · Dynamics

The fundamental principle of dynamics

Two lorries with identical engines, one empty and one loaded to the maximum, set off with the same engine force. The empty one shoots off; the loaded one takes much longer to pick up speed. The equation that explains this difference, as simple as $F = m \cdot a$, is the same one that makes it possible to calculate the trajectory of a rocket or the braking of a car.

Before you start, make sure you know: Changing units (2nd ESO)
Before you start, make sure you know: Acceleration (3rd ESO)

F = m · a: the fundamental principle

If you push an empty supermarket trolley, you can make it accelerate easily, but if it is loaded with bottles of water you need to push with far more force to get the same change in velocity: what exact mathematical relationship links the acceleration, the mass and the force you exert?

Block of mass m acted on by a resultant force to the right, producing an acceleration in the same sense, with the formula F = m · a below. m F resultant a F = m · a
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Calculate the net force required for a toy cart of $2\ \text{kg}$ to accelerate at $3\ \text{m/s}^2$.

Calculate the net force required for a toy cart of $2\ \text{kg}$ to accelerate at $3\ \text{m/s}^2$.

Solve
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A scooter together with its rider accelerates at $2\ \text{m/s}^2$ when a net force of $100\ \text{N}$ acts on the combination. Calculate the total mass.

A scooter together with its rider accelerates at $2\ \text{m/s}^2$ when a net force of $100\ \text{N}$ acts on the combination. Calculate the total mass.

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The newton: a definition based on the law itself

When you hold a medium-sized apple of about $100\ \text{g}$ in the palm of your hand, you feel a downward push due to Earth's gravity, but how do we formally define the standard international unit that allows us to measure any force precisely?

Body of mass 1 kg acted on by a force of 1 N producing an acceleration of 1 m/s², illustrating the definition of the newton. 1 kg 1 N 1 m/s² 1 N = 1 kg · 1 m/s²
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A net force of $20\ \text{N}$ acts on a box of $5\ \text{kg}$. What acceleration does the box acquire?

A net force of $20\ \text{N}$ acts on a box of $5\ \text{kg}$. What acceleration does the box acquire?

Solve
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Using Newton's second law, show that $1\ \text{N}$ is equivalent to $1\ \text{kg} \cdot \text{m/s}^2$ and give a simple numerical example to illustrate it.

Using Newton's second law, show that $1\ \text{N}$ is equivalent to $1\ \text{kg} \cdot \text{m/s}^2$ and give a simple numerical example to illustrate it.

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The resultant force: the vector sum of all the applied forces

Almost no body receives just one force. Before applying $F = m \cdot a$ you have to calculate the resultant force: the result of adding, as vectors, all the forces acting on the body, including friction if there is any.

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A person pushes a toolbox of $8\ \text{kg}$ with a force of $60\ \text{N}$, and the friction with the ground is $20\ \text{N}$. a) Calculate the net force on the box. b)…

A person pushes a toolbox of $8\ \text{kg}$ with a force of $60\ \text{N}$, and the friction with the ground is $20\ \text{N}$.

  1. a) Calculate the net force on the box.
  2. b) Calculate its acceleration.
Solve
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Three forces act on a crate of $10\ \text{kg}$ along the same line: $50\ \text{N}$ to the right, $30\ \text{N}$ to the right and $20\ \text{N}$ to the left. a) Calculate…

Three forces act on a crate of $10\ \text{kg}$ along the same line: $50\ \text{N}$ to the right, $30\ \text{N}$ to the right and $20\ \text{N}$ to the left.

  1. a) Calculate the net force.
  2. b) Calculate the acceleration of the crate.
  3. c) What additional force would need to be applied so that the crate moves at constant velocity? Reason your answer.
Solve
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A worker pushes a trolley of $20\ \text{kg}$ with a force of $100\ \text{N}$ and the friction is $20\ \text{N}$. a) Calculate the acceleration of the cart. b) If the…

A worker pushes a trolley of $20\ \text{kg}$ with a force of $100\ \text{N}$ and the friction is $20\ \text{N}$.

  1. a) Calculate the acceleration of the cart.
  2. b) If the worker doubles their push to $200\ \text{N}$ and the friction does not change, does the acceleration double as well? Calculate it and explain why this happens.
Solve

Constant velocity and balanced forces

A body moving does not mean it has a non-zero resultant force: if it travels at constant velocity (including being at rest), its acceleration is zero and, by the fundamental principle, so is the resultant force.

Lorry moving to the right at constant velocity: the engine force, 2000 N to the right, is balanced by a resistance of 2000 N to the left, so the resultant force and the acceleration are zero. lorry F engine = 2000 N resistance = 2000 N v v constant → F resultant = 0 → a = 0
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An aeroplane flies in a straight line at a constant velocity of $250\ \text{m/s}$. What is the net force on the aeroplane? What is its acceleration? Relate this to…

An aeroplane flies in a straight line at a constant velocity of $250\ \text{m/s}$. What is the net force on the aeroplane? What is its acceleration? Relate this to Newton's second law of motion.

Solve
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A sledge slides along horizontal ice and we stop pushing it. A classmate claims: "since there is no longer any force pushing it, it will stop straight away". a)…

A sledge slides along horizontal ice and we stop pushing it. A classmate claims: "since there is no longer any force pushing it, it will stop straight away".

  1. a) According to Newton's second law, how would the sledge move if friction were exactly zero? What would the net force on it be then?
  2. b) On real ice the sledge eventually comes to a stop: which force is responsible and in which direction does it act?
  3. c) Explain why the classmate's statement confuses "there is no applied force" with "there is no net force".
Solve

Mass and acceleration: inverse proportionality

Animation of two bodies starting from rest with the same resultant force: the 4000 kg one accelerates twice as much and covers twice the distance of the 8000 kg one, which lags behind, showing the inverse proportionality between mass and acceleration. same F 4000 kg large a 8000 kg small a
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A motorcycle with its rider has a mass of $200\ \text{kg}$ and travels at $72\ \text{km/h}$. a) Express the speed in $\text{m/s}$. b) Calculate the net braking force…

A motorcycle with its rider has a mass of $200\ \text{kg}$ and travels at $72\ \text{km/h}$.

  1. a) Express the speed in $\text{m/s}$.
  2. b) Calculate the net braking force required to stop in $4\ \text{s}$.
Solve
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In a laboratory, the acceleration of a trolley is measured when different forces are applied to it: with $2\ \text{N}$ it accelerates at $0,5\ \text{m/s}^2$; with…

In a laboratory, the acceleration of a trolley is measured when different forces are applied to it: with $2\ \text{N}$ it accelerates at $0,5\ \text{m/s}^2$; with $4\ \text{N}$, at $1\ \text{m/s}^2$; and with $6\ \text{N}$, at $1,5\ \text{m/s}^2$.

  1. a) Plot force against acceleration in your exercise book and describe the graph you obtain.
  2. b) Deduce the mass of the trolley from the data.
  3. c) Predict the acceleration of the trolley if a force of $10\ \text{N}$ were applied to it.
Solve
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Two lorries with identical engines, producing the same net force of $8000\ \text{N}$, start from rest: one is empty, with a mass of $4000\ \text{kg}$, and the other is…

Two lorries with identical engines, producing the same net force of $8000\ \text{N}$, start from rest: one is empty, with a mass of $4000\ \text{kg}$, and the other is loaded, with a mass of $8000\ \text{kg}$.

  1. a) Calculate the acceleration of each lorry.
  2. b) Calculate the time each lorry takes to reach $20\ \text{m/s}$.
  3. c) Using these numbers, explain why a loaded lorry needs more distance to join a motorway.
Solve

Summary

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