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Secondary 6 · Physics
Energy · Mechanical energy

Conservation of Mechanical Energy

A roller coaster car at the first peak: very high and almost motionless. The same car in the valley: low and moving incredibly fast. There is no motor between the two points, only gravity. On this page you will see when mechanical energy is conserved exactly, when it is dissipated, and how to use that difference to solve problems without needing the time of the motion.

Energy is transformed: potential ↔ kinetic

If you throw a ball upward at 10 m/s, it slows down until it stops at a height of 5 m, and when it returns to your hand it has a speed of 10 m/s again. Which quantity stayed constant throughout the whole trip?

A free-falling object in three positions: at the top it has maximum potential energy and zero kinetic energy, in the middle it has some of both, and at the bottom it has zero potential energy and maximum kinetic energy. top: Ep max, Ec = 0 middle: Ep and Ec bottom: Ep = 0, Ec max
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A stone of $2\ \text{kg}$ falls from a cliff of $20\ \text{m}$ (take $g = 10\ \text{m/s}^2$ and ignore air resistance). The situation is shown in the figure. a) What…

A stone of $2\ \text{kg}$ falls from a cliff of $20\ \text{m}$ (take $g = 10\ \text{m/s}^2$ and ignore air resistance). The situation is shown in the figure.

stone v = ? 20 m
  1. a) What potential energy does it have when it starts to fall?
  2. b) If mechanical energy is conserved, what kinetic energy will it have when it reaches the bottom?
  3. c) At what speed will it reach the bottom?
Solve
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You throw a ball of $0.5\ \text{kg}$ vertically upwards with a speed of $10\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$ and ignore friction), as shown in the figure. a)…

You throw a ball of $0.5\ \text{kg}$ vertically upwards with a speed of $10\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$ and ignore friction), as shown in the figure.

v₀ = 10 m/s h = ?
  1. a) What kinetic energy does it have when it leaves your hand?
  2. b) Apply the conservation of mechanical energy between the release point and the highest point (where the speed is zero) and calculate the maximum height reached.
Solve
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A skier goes down a slope, without pushing off, from a peak to the base (ignore friction), as shown in the figure. a) At which point of the descent is their speed…

A skier goes down a slope, without pushing off, from a peak to the base (ignore friction), as shown in the figure.

peak base v = ? h
  1. a) At which point of the descent is their speed maximum? Where is their potential energy maximum?
  2. b) Does their mechanical energy change during the descent?
  3. c) If another slope with a completely different profile connected the same peak to the same base, would they reach the bottom with a higher, lower, or equal speed? Justify your answer using conservation of energy.
Solve

Conservation with conservative forces

This explains why on a frictionless roller coaster the shape of the track does not matter: only the height counts. And that is why no later peak can be higher than the first one if the car starts almost from rest: it would not have the energy to climb higher.

A frictionless roller coaster car in two positions, one high and one lower, next to two energy bars: going down, the potential energy decreases and the kinetic energy increases, but the sum is the same. state 1 state 2 Em is conserved = Ep Ec Ep state 1 state 2

The loop: minimum speed at the top

Many roller coasters have a vertical loop. At the highest point of the loop, the car is "upside down" and the weight points toward the center of the circle. To follow the curve, the net force toward the center must be $m v^2 / r$ (centripetal force). If only the weight acted, we would have $m g = m v^2 / r$. With less speed, the car would come off the track and fall. Therefore, the minimum speed at the top of the loop is:

$$v_{min} = \sqrt{g\, r}$$

Why? If $v^2 < g r$, the weight alone would already be more force than is needed to turn, and the track cannot "pull" the car downward. If $v^2 > g r$, the track pushes the car downward with the extra force.

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On all roller coasters, the first peak of the track is the highest of the entire circuit. Look at the diagram: Explain, using the conservation of mechanical energy, why…

On all roller coasters, the first peak of the track is the highest of the entire circuit. Look at the diagram:

v₀ ≈ 0 h ?

Explain, using the conservation of mechanical energy, why no subsequent peak can exceed the height of the first peak if the carriage starts from it almost from rest, and what role friction plays in that conclusion.

Solve
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A roller coaster carriage passes the highest peak of the track, at $40\ \text{m}$, with a speed of $5\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$ and ignore friction). The…

A roller coaster carriage passes the highest peak of the track, at $40\ \text{m}$, with a speed of $5\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$ and ignore friction). The situation is shown in the figure.

5 m/s A B v = ? 40 m 10 m
  1. a) Calculate its mechanical energy per kilogram of mass.
  2. b) At what speed will it pass through a point located at a height of $10\ \text{m}$?
  3. c) Is it necessary to know the mass of the carriage to answer? Justify your answer using the conservation equation.
Solve
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A laboratory trolley enters the base of a curved track with a speed of $4\ \text{m/s}$ and travels up it without pushing off (take $g = 10\ \text{m/s}^2$ and ignore…

A laboratory trolley enters the base of a curved track with a speed of $4\ \text{m/s}$ and travels up it without pushing off (take $g = 10\ \text{m/s}^2$ and ignore friction).

  1. a) What maximum height does it reach?
  2. b) If it entered at twice the speed, $8\ \text{m/s}$, would it reach twice the height? Calculate it and comment on the result.
  3. c) Write down the proportional relationship between the height reached and the entry speed.
Solve
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A roller coaster carriage starts from rest at a peak $45\ \text{m}$ high, descends to a dip at $25\ \text{m}$ and then heads up a second peak of $40\ \text{m}$ (take…

A roller coaster carriage starts from rest at a peak $45\ \text{m}$ high, descends to a dip at $25\ \text{m}$ and then heads up a second peak of $40\ \text{m}$ (take $g = 10\ \text{m/s}^2$ and ignore friction). The profile of the track is shown in the figure.

? v = 0 v = ? v = ? 45 m 25 m 40 m 50 m
  1. a) What speed does it pass through the dip with?
  2. b) What speed does it pass over the second peak with?
  3. c) If there were a third peak at $50\ \text{m}$, would it manage to clear it? Justify your answer using energies and state the minimum height it should start from to achieve it (in the limit, arriving with zero speed).
  4. d) Taking into account that in reality there is always friction, explain whether the starting height should be greater or smaller than the one calculated in the previous section.
Solve
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A roller coaster carriage starts from rest at a peak of $45\ \text{m}$ and enters a loop of radius $12.5\ \text{m}$, whose highest point is therefore $25\ \text{m}$…

A roller coaster carriage starts from rest at a peak of $45\ \text{m}$ and enters a loop of radius $12.5\ \text{m}$, whose highest point is therefore $25\ \text{m}$ above the ground (take $g = 10\ \text{m/s}^2$ and ignore friction). The diagram is shown in the figure.

R = 12.5 m v = 0 v = ? v = ? 45 m 25 m
  1. a) At what speed does it pass through the lowest point of the circuit?
  2. b) At what speed does it pass through the highest point of the loop?
  3. c) For safety, the speed at the highest point of the loop must be at least $15\ \text{m/s}$ (given). Does the carriage exceed this? Calculate the minimum height from which the carriage could start in order to just satisfy this requirement.
  4. d) Explain why in the real design the starting peak is well above this theoretical minimum height.
Solve

Elastic energy: springs

Hooke's law says that a spring compressed or stretched a distance $x$ exerts a force $F = kx$. When it is deformed, it stores elastic potential energy $E_p = \frac12 k x^2$ (the ½ appears because the force grows from 0 to $kx$, and the work is the area under that graph: a triangle).

When the spring is released, that energy becomes kinetic energy of the object. On a frictionless table, mechanical energy is conserved: $\frac12 k x^2 = \frac12 m v^2$.

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A pogo stick has a spring that compresses on landing and stretches on take-off. a) Describe the sequence of energy transformations in a complete jump, from the moment…

A pogo stick has a spring that compresses on landing and stretches on take-off.

  1. a) Describe the sequence of energy transformations in a complete jump, from the moment the spring is fully compressed until the jumper reaches maximum height.
  2. b) Is mechanical energy conserved while the spring pushes? What about during flight? Reason both answers.
Solve
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A spring of $k = 200\ \text{N/m}$, compressed by $0.1\ \text{m}$, pushes a $0.5\ \text{kg}$ block initially at rest across a frictionless horizontal table. The diagram…

A spring of $k = 200\ \text{N/m}$, compressed by $0.1\ \text{m}$, pushes a $0.5\ \text{kg}$ block initially at rest across a frictionless horizontal table. The diagram is shown in the figure.

k = 200 N/m 0.5 kg x = 0.1 m v = ?
  1. a) What elastic potential energy does the compressed spring have?
  2. b) Apply the conservation of energy between the compressed spring and the block once separated from the spring, and calculate the speed at which it is launched.
  3. c) Explain what forces act on the block at each stage and why mechanical energy is conserved.
Solve
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A mass of $1\ \text{kg}$ hangs from a vertical spring of $k = 100\ \text{N/m}$. It is released from rest at the point where the spring is unstretched and falls,…

A mass of $1\ \text{kg}$ hangs from a vertical spring of $k = 100\ \text{N/m}$. It is released from rest at the point where the spring is unstretched and falls, stretching it (take $g = 10\ \text{m/s}^2$ and ignore friction). The diagram is shown in the figure.

k = 100 N/m 1 kg unstretched spring · v = 0 equilibrium position lowest point · v = 0 x = ? ?
  1. a) Explain why, both when released and at the lowest point of the fall, its kinetic energy is zero, and set up the conservation of energy between these two points: $m \times g \times x = \frac{1}{2} \times k \times x^2$, where $x$ is the maximum extension.
  2. b) Calculate the maximum extension $x$.
  3. c) The static equilibrium position (where the hanging mass would remain at rest) satisfies $k \times x_e = m \times g$: calculate it and compare it with $x$. Explain the factor $2$ that appears: where is the 'extra energy' when the mass passes through the equilibrium position?
Solve

Non-conservative forces: friction dissipates energy

After braking hard on a downhill road, the brakes and tires of the car are hot to the touch. Where did the kinetic energy the vehicle had go?

Energy balance with friction: the initial mechanical energy splits into final mechanical energy plus dissipated heat, equal to the absolute value of the work of friction. initial Em Before braking is split final Em heat = |W fric| After braking

Kinetic friction is $F_{fric} = \mu\, N$, where $N$ is the normal force. On a horizontal surface, $N = mg$.

Friction on an inclined plane

On an inclined plane the normal force is no longer equal to the weight. The weight $mg$ breaks down into a component parallel to the plane, $mg\sin\theta$, and a perpendicular one, $mg\cos\theta$. The plane only supports the perpendicular component, so:

$$N = m g \cos\theta, \qquad F_{fric} = \mu N = \mu\, m g \cos\theta$$

Why? The normal force balances the component of the weight that presses against the surface; the steeper the plane, the less it presses and the smaller the friction. If $\theta = 0$, $\cos\theta = 1$ and we recover $N = mg$.

The distance the object travels is the length of the plane, $L = h/\sin\theta$. The work of friction is calculated the same way and equals the change in mechanical energy:

$$W_{fric} = -F_{fric}\, L = \Delta E_m$$ Block on an inclined plane of height h and length L: the weight breaks down into a component parallel to the plane and a perpendicular one, which the normal force balances; friction points up the plane. h base L θ N friction weight mg
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A $2\ \text{kg}$ package slides down a metal ramp from a height of $5\ \text{m}$ and reaches the bottom with a speed of $8\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$).…

A $2\ \text{kg}$ package slides down a metal ramp from a height of $5\ \text{m}$ and reaches the bottom with a speed of $8\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$). The diagram is shown in the figure.

10 m package friction 5 m v = 8 m/s
  1. a) What potential energy did it have at the top?
  2. b) What kinetic energy does it have when it reaches the bottom?
  3. c) Has mechanical energy been conserved? Calculate the energy dissipated, which equals the magnitude of the work done by the friction force, $W_{roz} = \Delta E_m$.
  4. d) If the ramp is $10\ \text{m}$ long, calculate the average friction force acting on the package.
Solve
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A box of $4\ \text{kg}$ slides across a horizontal floor with an initial speed of $6\ \text{m/s}$ and comes to a stop due to friction. The coefficient of friction is…

A box of $4\ \text{kg}$ slides across a horizontal floor with an initial speed of $6\ \text{m/s}$ and comes to a stop due to friction. The coefficient of friction is $\mu = 0.3$ (take $g = 10\ \text{m/s}^2$).

  1. a) What kinetic energy did it have when it started to slide?
  2. b) Calculate the friction force, $F_{roz} = \mu \times m \times g$.
  3. c) The work done by friction dissipates all the kinetic energy: calculate the distance it travels before coming to a stop.
Solve
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A ball of $0.1\ \text{kg}$ is dropped from $2\ \text{m}$ and, after bouncing on the ground, only reaches a height of $1.5\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as…

A ball of $0.1\ \text{kg}$ is dropped from $2\ \text{m}$ and, after bouncing on the ground, only reaches a height of $1.5\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.

v = 0 2 m 1.5 m
  1. a) Calculate its potential energy before falling and at its highest point after the bounce.
  2. b) How much energy has it lost in the bounce? What percentage of the initial energy does this represent?
  3. c) What has this energy been transformed into?
  4. d) If the ball continued bouncing in the same way, what would eventually happen to its mechanical energy?
Solve
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The $0.2\ \text{kg}$ bob of a pendulum passes through the lowest point with a speed of $4\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$). a) What kinetic energy does it have…

The $0.2\ \text{kg}$ bob of a pendulum passes through the lowest point with a speed of $4\ \text{m/s}$ (take $g = 10\ \text{m/s}^2$).

  1. a) What kinetic energy does it have at that point?
  2. b) If there were no friction, what maximum height would it reach on the other side?
  3. c) In reality it only reaches $0.6\ \text{m}$: calculate the energy dissipated in this half-oscillation, $W_{roz} = \Delta E_m$.
  4. d) Explain, based on these numbers, why the pendulum will eventually stop and what its energy has been converted into.
Solve
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A spring of $k = 500\ \text{N/m}$, compressed by $0.2\ \text{m}$, launches a $1\ \text{kg}$ block across a horizontal floor with a coefficient of friction of…

A spring of $k = 500\ \text{N/m}$, compressed by $0.2\ \text{m}$, launches a $1\ \text{kg}$ block across a horizontal floor with a coefficient of friction of $\mu = 0.25$ (take $g = 10\ \text{m/s}^2$). The block travels a certain distance and comes to a stop.

  1. a) Calculate the initial elastic potential energy of the spring.
  2. b) Calculate the friction force acting on the block once it is moving.
  3. c) Explain why all the elastic energy is eventually dissipated and calculate the distance the block travels before stopping.
  4. d) If the spring were compressed by $0.4\ \text{m}$ (twice as much), what distance would it travel? Comment on the relationship you find.
Solve
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A car of $1200\ \text{kg}$ brakes to a complete stop on dry asphalt, with a coefficient of friction of $\mu = 0.6$ (take $g = 10\ \text{m/s}^2$). a) Calculate the…

A car of $1200\ \text{kg}$ brakes to a complete stop on dry asphalt, with a coefficient of friction of $\mu = 0.6$ (take $g = 10\ \text{m/s}^2$).

  1. a) Calculate the friction force during braking.
  2. b) If it was travelling at $30\ \text{m/s}$, calculate its kinetic energy and the braking distance (the work done by friction consumes all the kinetic energy).
  3. c) Repeat the calculation for an initial speed of $15\ \text{m/s}$.
  4. d) What is the relationship between the two braking distances? Explain the consequence of this result for urban speed limits.
Solve
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A block of $2\ \text{kg}$ slides down from rest along an inclined plane of height $3\ \text{m}$. The angle of the plane satisfies $\sen \theta = 0.6$ and…

A block of $2\ \text{kg}$ slides down from rest along an inclined plane of height $3\ \text{m}$. The angle of the plane satisfies $\sen \theta = 0.6$ and $\cos \theta = 0.8$, and the coefficient of friction is $\mu = 0.3$ (take $g = 10\ \text{m/s}^2$). The diagram is shown in the figure.

θ 3 m P N friction sin θ = 0.6 · cos θ = 0.8 · μ = 0.3
  1. a) Calculate the length of the plane covered by the block and the normal force acting on it, $N = m \times g \times \cos \theta$.
  2. b) Calculate the friction force and its work done, $W_{roz} = -F_{roz} \times L$.
  3. c) Apply $W_{roz} = \Delta E_m$ to calculate the kinetic energy of the block when reaching the bottom and its speed.
  4. d) What percentage of the initial potential energy has been dissipated by friction?
Solve
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A person and their equipment, totalling $80\ \text{kg}$, ride a zip line with $100\ \text{m}$ of cable that covers a height difference of $45\ \text{m}$. The average…

A person and their equipment, totalling $80\ \text{kg}$, ride a zip line with $100\ \text{m}$ of cable that covers a height difference of $45\ \text{m}$. The average friction (pulley, air resistance) is $200\ \text{N}$ (take $g = 10\ \text{m/s}^2$). The diagram is shown in the figure.

100 m 45 m v = ?
  1. a) What potential energy do they have at the start of the ride?
  2. b) Calculate the work done by friction during the entire descent, $W_{roz} = -F_{roz} \times d$.
  3. c) Calculate the kinetic energy and the speed with which they reach the end of the cable.
  4. d) At what speed would they arrive if there were no friction? Comment on the result and explain why zip lines need a braking system at the end.
Solve

Recognizing when mechanical energy is conserved

Situation Forces that do work Em?
Satellite in circular orbit Only gravity, perpendicular to the velocity Is conserved
Falling apple (no air) Only the weight Is conserved
Car braking hard Friction Decreases
Elevator rising at constant speed Motor (external force) Increases
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A swing swings freely after an initial push, as shown in the figure. a) At which points along the motion is its potential energy maximum? And where is its kinetic energy…

A swing swings freely after an initial push, as shown in the figure.

A B C
  1. a) At which points along the motion is its potential energy maximum? And where is its kinetic energy maximum?
  2. b) Describe the energy transformation that takes place as it descends from one extreme to the lowest point.
  3. c) If no one pushes it again, the amplitude gradually decreases: which forces are responsible, and what happens to the mechanical energy?
Solve
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State in which of these situations mechanical energy is conserved and in which it is not, justifying your answer with a sentence in each case: a) A satellite in a…

State in which of these situations mechanical energy is conserved and in which it is not, justifying your answer with a sentence in each case:

  1. a) A satellite in a circular orbit around the Earth.
  2. b) A car braking hard to a complete stop.
  3. c) An apple falling from a branch (ignore friction).
  4. d) A lift ascending at a constant speed.
Solve
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Explain in your own words what a conservative force is, using weight and the elastic force of a spring as examples, and what happens to the mechanical energy of a system…

Explain in your own words what a conservative force is, using weight and the elastic force of a spring as examples, and what happens to the mechanical energy of a system when only forces of this type act. Then give an example of a non-conservative force, state the sign of the work it does on a moving object, and explain its effect on the object's mechanical energy.

Solve
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In the laboratory, the $0.1\ \text{kg}$ bob of a pendulum is released from rest at a position $45\ \text{cm}$ higher than the lowest point of its swing (take…

In the laboratory, the $0.1\ \text{kg}$ bob of a pendulum is released from rest at a position $45\ \text{cm}$ higher than the lowest point of its swing (take $g = 10\ \text{m/s}^2$ and ignore friction), as shown in the figure.

v = 0 v = ? 45 cm
  1. a) Calculate its initial potential energy relative to the lowest point.
  2. b) Apply the conservation of energy and calculate the speed with which it passes through the lowest point.
  3. c) What height will it reach on the other side of the swing? Justify your answer without performing further calculations.
Solve
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A $2\ \text{kg}$ package drops from a stationary delivery drone at a height of $45\ \text{m}$ (take $g = 10\ \text{m/s}^2$ and ignore air resistance). a) What is its…

A $2\ \text{kg}$ package drops from a stationary delivery drone at a height of $45\ \text{m}$ (take $g = 10\ \text{m/s}^2$ and ignore air resistance).

  1. a) What is its mechanical energy when it starts to fall?
  2. b) As it passes a height of $25\ \text{m}$, how much potential energy and how much kinetic energy does it have? At what speed does it pass through there?
  3. c) At what speed does it reach the ground?
  4. d) Do the calculated speeds depend on the mass of the package? Justify your answer using the conservation equation.
Solve

Back to the beginning: in the valley of the roller coaster all the potential energy from the peak has been transformed into kinetic energy; since $m g h = \frac12 m v^2$, the speed is $v = \sqrt{2gh}$: it depends on the height, not on the mass of the car or the shape of the track.

Summary

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