In mechanics, define the work done by a force.

Work is done by a force when that force causes the object to move: both a force and movement are needed. A force that merely holds an object stationary does no work at all, however large it is.

An object is pushed up an inclined plane. Which of the forces acting on it does no work, and why?

The normal reaction . Work counts only the component of a force in the direction of motion , and the normal reaction acts perpendicular to the plane, so it has no component along the slope at all.

A box is dragged along a rough floor at constant speed . What does the constant speed tell you, and how much work is done against friction?

Constant speed means zero acceleration, so the forces along the direction of motion balance : the forward component of the pull equals the frictional force. The work done against friction is then that frictional force multiplied by the distance moved.

In this course, what is meant by mechanical energy ?

Mechanical energy is the kind of energy dealt with in mechanics, as distinct from heat, light, chemical or nuclear energy. It exists in two forms: kinetic energy (KE) and gravitational potential energy (GPE) .

In the work-energy principle, when do you add the work done and when do you subtract it?

Add it for forces that help the object move forwards, such as a driving force or a tension pulling it along. Subtract it for forces that hinder the motion, such as friction, air resistance or a tension pulling backwards. Where several such forces act, add or subtract each one according to which it does.

Give the two ways a situation can have no work done by non-gravitational forces.

There may be none acting , as for a particle falling freely under gravity from a fixed height. Or every one of them may act perpendicular to the direction of motion , as the normal reaction does on an object moving along a horizontal surface.

International A Level (IAL)MathsEdexcelMechanics 2FlashcardsWork, Energy & PowerWork & Energy

Work & Energy (Edexcel International A Level (IAL) Maths: Mechanics 2): Flashcards

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In mechanics, define the work done by a force.

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In mechanics, define the work done by a force.
Work is done by a force when that force causes the object to move: both a force and movement are needed.
A force that merely holds an object stationary does no work at all, however large it is.
A force of magnitude $F$ N acts at an angle $θ$ to the direction of motion while the object moves a distance $d$ m. Fill in the blank to complete the work done:
$W = F d______$
The completed formula is:
$W = F d \cos θ$
Only the component of the force in the direction of motion counts, and that component is $F \cos θ$ .
An object is pushed up an inclined plane. Which of the forces acting on it does no work, and why?
The normal reaction.
Work counts only the component of a force in the direction of motion, and the normal reaction acts perpendicular to the plane, so it has no component along the slope at all.
True or False?
1 joule is the work done by a force of 1 newton moving an object 1 metre.
True, provided the metre is measured along the line of action of the force.
That is also why work can be measured in newton metres, and why $1 J = 1 N m$ .
Larger amounts are given in kilojoules, with $1 kJ = 1000 J$ .
A box is dragged along a rough floor at constant speed. What does the constant speed tell you, and how much work is done against friction?
Constant speed means zero acceleration, so the forces along the direction of motion balance: the forward component of the pull equals the frictional force.
The work done against friction is then that frictional force multiplied by the distance moved.
A child pulls a box 5 m along a rough floor at constant speed with a force of 10 N at $30 °$ to the horizontal. Find the work done against friction.
At constant speed the friction equals the horizontal component of the pull, so the work done against friction is the work done by that component:
$W = 10 \times 5 \times \cos 30 ° = 43.3 J$
The mass of the box plays no part, because the vertical direction is never used.
In this course, what is meant by mechanical energy?
Mechanical energy is the kind of energy dealt with in mechanics, as distinct from heat, light, chemical or nuclear energy.
It exists in two forms: kinetic energy (KE) and gravitational potential energy (GPE).
A particle of mass $m$ kg moves with speed $v {m s}^{- 1}$ . Fill in the blanks to complete its kinetic energy:
$KE =______m v^{______}$
The completed formula is:
$KE = \frac{1}{2} m v^{2}$
A particle has kinetic energy only while it is moving, and this is one of the formulae you are expected to know rather than look up.
Kinetic energy is a scalar. What does that tell you about its sign, and what unit is it measured in?
It can never be negative, which the formula guarantees, since $v^{2}$ is never negative whichever way the particle is travelling.
It is measured in joules (J), with $1 kJ = 1000 J$ .
How do you find the change in kinetic energy, and what is the common mistake?
Subtract the initial kinetic energy from the final one:
$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = \frac{1}{2} m (v^{2} - u^{2})$
The mistake is to use the difference of the speeds where the difference of their squares is wanted: $v^{2} - u^{2}$ is not ${(v - u)}^{2}$ .
A particle moves in two dimensions with velocity vector $v$ . What are the two ways to find its kinetic energy?
Find the speed as the magnitude of $v$ , then put that single number into the formula.
Or apply the formula to each component separately and add the two results together.
A jogger's speed rises from $2 {m s}^{- 1}$ to $3 {m s}^{- 1}$ and her kinetic energy increases by 150 J. Find her mass.
Use the difference of the squares of the speeds:
$\frac{1}{2} m (3^{2} - 2^{2}) = 150$
That gives $\frac{5}{2} m = 150$ , so the jogger's mass is $60 kg$ .
What is gravitational potential energy, and what is the formula for it?
Gravitational potential energy is the energy an object has because of its height above a chosen fixed level, with gravity acting on it.
It is the weight multiplied by that height, $GPE = m g h$ , measured in joules when $m$ is in kilograms and $h$ in metres.
An object is dragged up a slope. Which height goes into $m g h$ , and what if you are given the slant distance?
The vertical height gained, never the distance travelled along the slope.
If the question gives you the slant distance, use trigonometry on the angle of the slope to find the vertical rise first.
True or False?
An object resting on the chosen base level has zero gravitational potential energy.
True.
Height is measured from whatever fixed level has been chosen as the base, so $h = 0$ there and $m g h = 0$ .
That also means gravitational potential energy is not an absolute quantity: move the base level and every value changes, which is why problems turn on changes in it.
Fill in the blanks to complete the work-energy principle, written as an energy balance:
total final energy $=$ total $______$ energy $\pm$ work done by $______$ forces.
The completed principle is:
total final energy $=$ total initial energy $\pm$ work done by non-gravitational forces.
It is an energy balance, rather like a bank account: the final amount is the initial amount plus whatever was put in or taken out.
In the work-energy principle, what does "total energy" mean?
The sum of the gravitational potential energy and the kinetic energy, worked out at each of the two instants.
So each side of the balance is of the form $m g h + \frac{1}{2} m v^{2}$ , one with the initial values and one with the final ones.
Why is the weight $m g$ left out of the "work done by non-gravitational forces" term?
Because the work done against gravity has already been counted, inside the gravitational potential energy part of the total energy.
Putting the weight into the work-done term as well would count the same energy twice.
The forces that do belong there are friction, tensions, driving forces and air resistance.
In the work-energy principle, when do you add the work done and when do you subtract it?
Add it for forces that help the object move forwards, such as a driving force or a tension pulling it along.
Subtract it for forces that hinder the motion, such as friction, air resistance or a tension pulling backwards.
Where several such forces act, add or subtract each one according to which it does.
True or False?
The "change in KE $=$ minus change in GPE $\pm$ work done" form works in every situation without adaptation.
False.
That gain-and-loss form has to be adapted to each situation, and it is where most sign errors come from, because a "gain" is sometimes negative.
The energy-balance form, final $=$ initial $\pm$ work done, is written the same way every time and works in every situation.
Define conservation of energy.
Conservation of energy is the special case of the work-energy principle in which no work is done by non-gravitational forces.
The balance then reduces to total final energy $=$ total initial energy, with no extra term at all.
Give the two ways a situation can have no work done by non-gravitational forces.
There may be none acting, as for a particle falling freely under gravity from a fixed height.
Or every one of them may act perpendicular to the direction of motion, as the normal reaction does on an object moving along a horizontal surface.
A particle is fired horizontally at $8 {m s}^{- 1}$ off a 30 m cliff and moves freely under gravity. Use energy to find its speed on impact.
Nothing but gravity acts, so the total energy is unchanged and the gravitational potential energy lost becomes kinetic energy gained:
$\frac{1}{2} m v^{2} = \frac{1}{2} m \times 8^{2} + m g \times 30$
The mass cancels throughout, leaving $v^{2} = 64 + 588 = 652$ , so the speed of impact is $25.5 {m s}^{- 1}$ to 3 significant figures.

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