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ESO Y4A · Physics and Chemistry
Energy · Work and mechanical energy

Work and power

In 3rd ESO you saw that a force does work when it manages to move something in its own direction: $W = F \cdot d$. But forces do not always pull in a straight line with the motion. A child dragging a cart with the rope at an angle, a lawnmower pushed by a handle held diagonally: here the force and the displacement form an angle, and the formula needs refining.

Before you start, make sure you know: Work and power (3rd ESO)
Before you start, make sure you know: First-degree equations (2nd ESO)

Revision: when the force and the displacement are in the same line

If you push with all your might against a heavy wardrobe but cannot move it from the spot, you end up exhausted, yet in physics we say you have not transferred any energy to the wardrobe: why does physics say there is no work if there is no displacement?

A body on a horizontal surface is pushed by a force F in the same direction and sense as its displacement d, with the expression W = F · d body F d W = F · d
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A worker pushes a pallet with a horizontal force of $150\ \text{N}$ and moves it $8\ \text{m}$ along the warehouse floor. Calculate the work done.

A worker pushes a pallet with a horizontal force of $150\ \text{N}$ and moves it $8\ \text{m}$ along the warehouse floor. Calculate the work done.

Solve
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You place a box of books of $8\ \text{kg}$ on the top shelf of a cupboard, $1,5\ \text{m}$ above the floor (take $g = 10\ \text{m/s}^2$). a) What minimum force do you…

You place a box of books of $8\ \text{kg}$ on the top shelf of a cupboard, $1,5\ \text{m}$ above the floor (take $g = 10\ \text{m/s}^2$).

  1. a) What minimum force do you have to overcome to raise it?
  2. b) What work do you do against gravity?
Solve

When the force is at an angle: the cosine comes into play

When you pull a wheeled suitcase, if you hold the handle almost horizontal it moves along easily, but if you raise it almost vertical it is much harder to make it move forwards: how much of your force is really used depending on the angle?

Animation of the force F, of constant length, whose angle alpha with the horizontal displacement turns from about 33° to 90°, then to 0° and back to the start, while its dashed horizontal projection (the useful part, F·cos α) gets shorter and longer. F α d W = F · d · cos α
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A child pulls a cart with a rope that forms an angle of $60^\circ$ with the horizontal, exerting a force of $200\ \text{N}$, and drags it $10\ \text{m}$ (…

A child pulls a cart with a rope that forms an angle of $60^\circ$ with the horizontal, exerting a force of $200\ \text{N}$, and drags it $10\ \text{m}$ ($\cos 60^\circ = 0,5$), as shown in the figure.

F = 200 N 60° d = 10 m W = ?

Calculate the work done.

Solve
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Aitana mows the lawn by pushing the lawnmower with a force of $80\ \text{N}$ applied to the handle, which forms an angle of $30^\circ$ with the horizontal (…

Aitana mows the lawn by pushing the lawnmower with a force of $80\ \text{N}$ applied to the handle, which forms an angle of $30^\circ$ with the horizontal ($\cos 30^\circ \approx 0,87$), and covers $20\ \text{m}$ of lawn, as shown in the figure.

F = 80 N 30° Fₓ d = 20 m W = ?
  1. a) Calculate the work done.
  2. b) What work would have been done by pushing horizontally with the same force? What accounts for the difference?
Solve

Special cases: 0°, 90° and 180°

Comparison of the work depending on the angle between force and displacement: positive for 0°, zero for 90° and negative for 180° α = 0°, W > 0 force and displacement in the same sense α = 180°, W < 0 force in the opposite sense to the motion α = 90°, W = 0
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The work done by a force on a moving object is calculated as $W = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. You…

The work done by a force on a moving object is calculated as $W = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. You pull a trolley with the same force and over the same distance in three cases: with the rope horizontal ($\alpha = 0^\circ$), at an angle of $45^\circ$, and at an angle of $90^\circ$ ($\cos 0^\circ = 1$; $\cos 45^\circ \approx 0,71$; $\cos 90^\circ = 0$). The figure shows the three cases.

F α = 0° F α = 45° F α = 90°

Order the three amounts of work done from greatest to least and justify your answer using the formula.

Solve
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A worker pushes a container $12\ \text{m}$ along a warehouse floor with a horizontal force of $80\ \text{N}$, while friction opposes it with $45\ \text{N}$, as shown in…

A worker pushes a container $12\ \text{m}$ along a warehouse floor with a horizontal force of $80\ \text{N}$, while friction opposes it with $45\ \text{N}$, as shown in the figure. Remember that work done by a force opposing displacement is negative ($\alpha = 180^\circ$, $\cos 180^\circ = -1$).

container F = 80 N friction = 45 N d = 12 m W = ?
  1. a) Calculate the work done by the worker.
  2. b) Calculate the work done by friction.
  3. c) What is the net work done on the container?
Solve

Power: the rate of transferring energy

A crane with a motor lifts a box to a certain height while a stopwatch measures the time; the work W in joules divided by the time t in seconds gives the power P in watts P (W) W (J) t (s) W (J) ÷ t (s) = P (W)
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The motor of a conveyor belt in a parcel distribution centre does $18000\ \text{J}$ of work in $60\ \text{s}$. Power is calculated as $P = W/t$. Calculate the power of…

The motor of a conveyor belt in a parcel distribution centre does $18000\ \text{J}$ of work in $60\ \text{s}$. Power is calculated as $P = W/t$. Calculate the power of the motor, in watts.

Solve
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A drill has a power rating of $600\ \text{W}$ and operates for $2\ \text{min}$ ($1\ \text{kWh} = 3,6\cdot10^{6}\ \text{J}$). a) Express the time in seconds. b) Calculate…

A drill has a power rating of $600\ \text{W}$ and operates for $2\ \text{min}$ ($1\ \text{kWh} = 3,6\cdot10^{6}\ \text{J}$).

  1. a) Express the time in seconds.
  2. b) Calculate the work (energy) done.
  3. c) Express this energy in $\text{kWh}$.
Solve
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The motor of a patio awning has a power rating of $40\ \text{W}$ and takes $15\ \text{s}$ to retract it. a) What work does it do? b) If the awning rises by…

The motor of a patio awning has a power rating of $40\ \text{W}$ and takes $15\ \text{s}$ to retract it.

  1. a) What work does it do?
  2. b) If the awning rises by $1,5\ \text{m}$, what average force does the motor exert?
Solve

The same work by different routes

Comparison between lifting a load directly vertically or lifting it along an inclined ramp vertical ramp
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A kayaker drags his kayak $40\ \text{m}$ across the sand by pulling a rope that forms an angle of $45^\circ$ with the horizontal, with a force of $50\ \text{N}$ (…

A kayaker drags his kayak $40\ \text{m}$ across the sand by pulling a rope that forms an angle of $45^\circ$ with the horizontal, with a force of $50\ \text{N}$ ($\cos 45^\circ \approx 0,71$). Friction with the sand opposes the motion with $20\ \text{N}$.

  1. a) Calculate the work done by the kayaker.
  2. b) Calculate the work done by friction (remember it is negative).
  3. c) What is the net work done on the kayak? How can you observe the effect of this net work?
Solve
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To load a box of $60\ \text{kg}$ onto a lorry, whose bed is $1,2\ \text{m}$ above the ground, you can lift it vertically or push it up a frictionless ramp $3\ \text{m}$…

To load a box of $60\ \text{kg}$ onto a lorry, whose bed is $1,2\ \text{m}$ above the ground, you can lift it vertically or push it up a frictionless ramp $3\ \text{m}$ long (take $g = 10\ \text{m/s}^2$). The figure shows the two options.

lorry 1,2 m F vertically F 3 m ramp W = ?
  1. a) Calculate the work required in each case and explain why it is the same.
  2. b) Calculate the force that must be applied in each case.
  3. c) What is the advantage of using the ramp? What is the "cost" of using it?
Solve

Back to the beginning: with the lawnmower, pushing with the handle at an angle means that only the horizontal component of your force ($F \cdot \cos \alpha$) does work to move it forwards; the vertical component just presses the machine into the grass.

Summary

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