Work and power
In 3rd ESO you saw that a force does work when it manages to move an object in the same direction as the force. In the academic pathway you are going to treat work as what it really is: the scalar product of two vectors, force and displacement. This explains once and for all why the work is sometimes less than you would expect, why it can be negative, and how to relate the instantaneous force of a motor to the speed at which it travels.
Work as a scalar product: $W = \vec F \cdot \vec d$
You pull a suitcase with a constant force of $40\ \text{N}$: if you pull horizontally it takes less effort to travel the same distance than if you pull the strap at an angle of $45^\circ$, where you notice that part of the effort is wasted pulling upwards: how much energy do you actually make use of when you tilt the force?
It helps to think of it in terms of components: of the whole force, only the component in the direction of the displacement "counts" towards the work, $F_x = F \cos \alpha$. The perpendicular component, $F \sin \alpha$, does not move anything in that direction and therefore contributes no work.
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.
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.
Calculate the work done.
The three limiting cases and the work done by friction
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.
Order the three amounts of work done from greatest to least and justify your answer using the formula.
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.
- a) Calculate the work done.
- b) What work would have been done by pushing horizontally with the same force? What accounts for the difference?
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 $20\ \text{N}$.
- a) Calculate the work done by the kayaker.
- b) Calculate the work done by friction (remember it is negative).
- c) What is the net work done on the kayak? How can you observe the effect of this net work?
Net work and the change in kinetic energy
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 ($\alpha = 180^\circ$, $\cos 180^\circ = -1$).
- a) Calculate the work done by the worker.
- b) Calculate the work done by friction.
- c) What is the net work done on the container?
In a weightlifting competition, an athlete lifts a weight of $100\ \text{kg}$ from the floor to a height of $2\ \text{m}$ in $2\ \text{s}$, holds it still above her head…
In a weightlifting competition, an athlete lifts a weight of $100\ \text{kg}$ from the floor to a height of $2\ \text{m}$ in $2\ \text{s}$, holds it still above her head for $3\ \text{s}$, and then walks $10\ \text{m}$ horizontally with it until leaving the platform (take $g = 10\ \text{m/s}^2$). The figure shows the three phases.
- a) What work does she do against gravity while lifting it?
- b) What power does she develop during this lift?
- c) Does she do work against gravity while holding it still? What about while walking horizontally? Justify both answers using the definition of work.
Power: the rate of transferring energy
Two hoists lift a pallet of bricks of mass $400\ \text{kg}$ up to the third floor ($10\ \text{m}$): both do exactly the same work ($40\,000\ \text{J}$), but model A takes $8\ \text{s}$ and model B takes $40\ \text{s}$: how do we measure which of the two transfers energy at the greater rate?
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.
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 $2\ \text{min}$ ($1\ \text{kWh} = 3,6\cdot10^{6}\ \text{J}$).
- a) Express the time in seconds.
- b) Calculate the work (energy) done.
- c) Express this energy in $\text{kWh}$.
The human heart does about $1,2\ \text{J}$ of work with each beat ($1\ \text{kWh} = 3,6\cdot10^{6}\ \text{J}$). a) Calculate its average power at rest, with $60$ beats…
The human heart does about $1,2\ \text{J}$ of work with each beat ($1\ \text{kWh} = 3,6\cdot10^{6}\ \text{J}$).
- a) Calculate its average power at rest, with $60$ beats per minute.
- b) How much energy does it pump in a full day? Express it in joules and in $\text{kWh}$.
- c) Compare it with the power of a $10\ \text{W}$ LED light bulb and comment on the result.
Useful power and efficiency in real machines
A 4x4 car of $1800\ \text{kg}$ drives up a mountain pass gaining $100\ \text{m}$ in altitude in $2\ \text{min}$, at constant speed (take $g = 10\ \text{m/s}^2$). a)…
A 4x4 car of $1800\ \text{kg}$ drives up a mountain pass gaining $100\ \text{m}$ in altitude in $2\ \text{min}$, at constant speed (take $g = 10\ \text{m/s}^2$).
- a) Calculate the work done against gravity.
- b) Calculate the useful power used to climb, in $\text{kW}$.
- c) The car's engine specifies a power rating of $150\ \text{kW}$: explain at least two reasons why the useful power you calculated is much lower.
In an automated warehouse, a robot lifts packages of $5\ \text{kg}$ onto a shelf $6\ \text{m}$ high, at a rate of $120$ packages per hour (take $g = 10\ \text{m/s}^2$).…
In an automated warehouse, a robot lifts packages of $5\ \text{kg}$ onto a shelf $6\ \text{m}$ high, at a rate of $120$ packages per hour (take $g = 10\ \text{m/s}^2$).
- a) What work does it do with each package?
- b) What total work does it do in one hour?
- c) What average power does it develop?
- d) The robot's motor is rated at $500\ \text{W}$, considerably more than what you calculated: give reasons why this margin is needed.
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.
- a) Calculate the work required in each case and explain why it is the same.
- b) Calculate the force that must be applied in each case.
- c) What is the advantage of using the ramp? What is the "cost" of using it?