True or False? A person holding a heavy suitcase still is doing work on it.

False. Mechanical work needs a force and movement in the direction of that force, and the suitcase does not move. Holding it may be tiring, but no mechanical work is done on the suitcase at all.

True or False? A force can act on a moving object and do no work on it.

True. If the force acts perpendicular to the direction of motion, its component along the motion is zero, so it does no work however large it is. The normal reaction on an object sliding along a surface is the standard example.

When does work done on an object add energy, and when does it remove energy?

Work done by a force acting in the direction of motion adds energy to the object. Work done against a resistive force such as friction removes energy, and the net effect on the object is the difference between the two.

Does an object's gravitational potential energy have a single fixed value?

No: it depends on the level you choose to measure heights from, and that choice is yours to make. Only changes in gravitational potential energy matter in a problem, and those come out the same whichever base level you pick.

A particle is lifted vertically. How is the work done against gravity related to its potential energy?

The work done against gravity in lifting it is equal to the gain in gravitational potential energy. Coming back down, gravity does that same amount of work on the particle and its potential energy falls by the same amount, so the two are two descriptions of one transfer.

Why is the weight left out of the work done term in the work-energy principle?

Because the effect of gravity is already counted in the gravitational potential energy terms. Adding the work done against gravity as well would count the same transfer of energy twice over, which is why only non-gravitational forces appear in that term.

Why can an energy method handle a curved path where the suvat formulae cannot?

Because the energy balance compares only the initial and final states, and does not care what happened in between. The constant-acceleration formulae need one constant acceleration along a straight line, which a curved path does not provide.

True or False? On a rough slope the sum of a particle's kinetic and potential energies stays the same.

False. Friction is a non-gravitational force doing work against the motion, so it removes energy from that sum. The kinetic and potential energies at the end add to less than they did at the start, by exactly the work done against friction.

A LevelMathsCambridge (CIE)MechanicsFlashcardsEnergy, Work & PowerWork & Energy

Work & Energy (Cambridge (CIE) A Level Maths: Mechanics): Flashcards

Exam code: 9709

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Define work done.

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Cards in this collection (24)

Define work done.
The work done by a force is the force multiplied by the distance its point of application moves in the direction of the force.
For a constant force $F$ moving an object a distance $d$ along its own line of action,
$W = F d$
How much work is done by a force acting at an angle to the direction of motion?
Only the component along the direction of motion does any work, so
$W = F d \cos θ$
Here $θ$ is the angle between the force and the direction in which the object actually moves.
True or False?
A person holding a heavy suitcase still is doing work on it.
False.
Mechanical work needs a force and movement in the direction of that force, and the suitcase does not move.
Holding it may be tiring, but no mechanical work is done on the suitcase at all.
True or False?
A force can act on a moving object and do no work on it.
True.
If the force acts perpendicular to the direction of motion, its component along the motion is zero, so it does no work however large it is.
The normal reaction on an object sliding along a surface is the standard example.
Fill in the unit of work done and what it is equivalent to:
Work done is measured in $______$ , and one of them is the same as one newton $______$ .
Work done is measured in joules, and one of them is the same as one newton metre.
One joule is the work done when a force of one newton moves an object one metre along its line of action, and $1 kJ = 1000 J$ .
A box is pulled $5 m$ along a rough floor at constant speed by a force of $10 N$ at $30 °$ to the horizontal. Find the work done against friction.
At constant speed the horizontal forces balance, so the friction is equal to the horizontal component $10 \cos 30 °$ .
The work done against friction is therefore $10 \cos 30 ° \times 5 = 43.3 J$ to three significant figures.
A crate is pushed $6 m$ up a ramp inclined at $15 °$ . Which distance is used to find the work done against gravity?
The vertical height gained, not the distance travelled along the ramp: here $h = 6 \sin 15 ° = 1.55 m$ .
Gravity acts vertically, so only vertical movement counts against it, and using the $6 m$ would give an answer far too large.
When does work done on an object add energy, and when does it remove energy?
Work done by a force acting in the direction of motion adds energy to the object.
Work done against a resistive force such as friction removes energy, and the net effect on the object is the difference between the two.
Define kinetic energy.
Kinetic energy is the energy an object has because of its motion, given by
$KE = \frac{1}{2} m v^{2}$
It is measured in joules, and a particle has kinetic energy only while it is moving.
A particle's speed changes from $u$ to $v$ . Complete the expression for its change in kinetic energy:
$change in KE = \frac{1}{2} m______- \frac{1}{2} m______$
The completed expression is:
$change in KE = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$
This is the difference of the squares of the two speeds, not the square of the difference, which is the commonest slip on this topic.
How does $F = m a$ with $v^{2} = u^{2} + 2 a d$ give the change in kinetic energy?
Rearranging the second gives $a = \frac{v^{2} - u^{2}}{2 d}$ , and substituting it into the first gives $F = \frac{m (v^{2} - u^{2})}{2 d}$ .
Multiplying both sides by $d$ leaves $F d = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ , so the work done by the resultant force is the change in kinetic energy.
True or False?
Kinetic energy is negative when an object is moving backwards.
False.
Kinetic energy is a scalar and depends on $v^{2}$ , which is positive whichever way the object is travelling.
A particle moving backwards at $3 {m s}^{- 1}$ has exactly the same kinetic energy as one moving forwards at $3 {m s}^{- 1}$ .
Define gravitational potential energy.
Gravitational potential energy is the energy an object has because of its height above a chosen level, given by
$GPE = m g h$
It is measured in joules, and an object sitting at the chosen base level has none.
Does an object's gravitational potential energy have a single fixed value?
No: it depends on the level you choose to measure heights from, and that choice is yours to make.
Only changes in gravitational potential energy matter in a problem, and those come out the same whichever base level you pick.
A particle is lifted vertically. How is the work done against gravity related to its potential energy?
The work done against gravity in lifting it is equal to the gain in gravitational potential energy.
Coming back down, gravity does that same amount of work on the particle and its potential energy falls by the same amount, so the two are two descriptions of one transfer.
A ball of mass $400 g$ rises from $1 m$ to $4 m$ above the ground. Taking $g = 10 {m s}^{- 2}$ , find the change in its potential energy.
The mass is $0.4 kg$ and the height gained is $4 - 1 = 3 m$ .
So the gain is $m g h = 0.4 \times 10 \times 3 = 12 J$ , and it is a gain because the ball has risen.
State the work-energy principle.
The total final energy equals the total initial energy, plus or minus the work done by non-gravitational forces:
$E_{f} = E_{i} \pm WD$
Here the total energy means the gravitational potential energy and the kinetic energy added together.
Why is the weight left out of the work done term in the work-energy principle?
Because the effect of gravity is already counted in the gravitational potential energy terms.
Adding the work done against gravity as well would count the same transfer of energy twice over, which is why only non-gravitational forces appear in that term.
Fill in the two signs used for work done in the work-energy principle:
work done by a force that helps the object move forwards takes a $______$ sign
work done by a force that resists the motion takes a $______$ sign
Work done by a force that helps the object move forwards takes a plus sign.
Work done by a force that resists the motion takes a minus sign.
A driving force or a forward tension helps; friction, air resistance and a backward tension resist.
A toy of mass $0.5 kg$ is pushed $10 m$ up a rough $20 °$ slope by a force $P$ parallel to the slope, starting from rest and reaching $1 {m s}^{- 1}$ , with $μ = 0.1$ and $g = 10 {m s}^{- 2}$ . Find $P$ .
The gain in potential energy is $0.5 \times 10 \times 10 \sin 20 ° = 17.1 J$ and the gain in kinetic energy is $0.25 J$ .
The friction is $0.1 \times 0.5 \times 10 \cos 20 ° = 0.470 N$ , so it takes $4.70 J$ of work over the $10 m$ .
Balancing gives $17.1 + 0.25 = 10 P - 4.70$ , so $P = 2.20 N$ .
Why can an energy method handle a curved path where the suvat formulae cannot?
Because the energy balance compares only the initial and final states, and does not care what happened in between.
The constant-acceleration formulae need one constant acceleration along a straight line, which a curved path does not provide.
Define conservation of energy, as used in mechanics.
Conservation of energy is the special case of the work-energy principle in which no work is done by non-gravitational forces, so
$m g h_{f} + \frac{1}{2} m v_{f}^{2} = m g h_{i} + \frac{1}{2} m v_{i}^{2}$
The potential and kinetic energies then add to the same total at every point of the motion.
A particle is fired horizontally at $8 {m s}^{- 1}$ from a $30 m$ cliff and falls freely. Taking $g = 10 {m s}^{- 2}$ , find its speed on impact.
The potential energy lost equals the kinetic energy gained, so $\frac{1}{2} v^{2} = \frac{1}{2} (8^{2}) + 10 \times 30$ .
That gives $v^{2} = 64 + 600 = 664$ , so $v = 25.8 {m s}^{- 1}$ to three significant figures.
The mass cancels from every term, so it is never needed.
True or False?
On a rough slope the sum of a particle's kinetic and potential energies stays the same.
False.
Friction is a non-gravitational force doing work against the motion, so it removes energy from that sum.
The kinetic and potential energies at the end add to less than they did at the start, by exactly the work done against friction.

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