Conservation in conservative and non-conservative systems
A roller-coaster car at its first peak, very high up and almost at rest; the same car in the valley, low down and moving very fast. Between the two points there is no engine and there are no brakes: only gravity. In this section you will formalise when mechanical energy is conserved exactly and when it is not, and use that difference to solve problems involving falls, springs and braking without ever needing to know how long the motion lasts.
Energy is transformed: potential ↔ kinetic
If you throw a ball upwards at 10 m/s, it loses speed until it stops at a height of 5 m and, when it falls back into your hand, it again reads exactly 10 m/s: which physical quantity has remained constant throughout the whole journey?
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) At which points along the motion is its potential energy maximum? And where is its kinetic energy maximum?
- b) Describe the energy transformation that takes place as it descends from one extreme to the lowest point.
- c) If no one pushes it again, the amplitude gradually decreases: which forces are responsible, and what happens to the mechanical energy?
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.
- a) What kinetic energy does it have when it leaves your hand?
- 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.
Conservation of mechanical energy with conservative forces
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.
- a) What potential energy does it have when it starts to fall?
- b) If mechanical energy is conserved, what kinetic energy will it have when it reaches the bottom?
- c) At what speed will it reach the bottom?
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.
- a) Calculate its mechanical energy per kilogram of mass.
- b) At what speed will it pass through a point located at a height of $10\ \text{m}$?
- c) Is it necessary to know the mass of the carriage to answer? Justify your answer using the conservation equation.
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.
- a) What speed does it pass through the dip with?
- b) What speed does it pass over the second peak with?
- 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).
- 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.
Elastic energy and friction combined
In a toy launcher, compressing the spring by 1 cm barely moves the projectile half a metre across the carpet, but compressing it by 2 cm sends it a full two metres: why does the distance travelled grow so much faster than the deformation applied?
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.
- a) Calculate the initial elastic potential energy of the spring.
- b) Calculate the friction force acting on the block once it is moving.
- c) Explain why all the elastic energy is eventually dissipated and calculate the distance the block travels before stopping.
- 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.
Non-conservative forces: friction dissipates mechanical energy
After braking hard on a steep downhill road, the brake discs and tyres are very hot to the touch: where has the kinetic energy the vehicle had before braking gone?
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 $g = 10\ \text{m/s}^2$).
- a) Calculate the friction force during braking.
- 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).
- c) Repeat the calculation for an initial speed of $15\ \text{m/s}$.
- d) What is the relationship between the two braking distances? Explain the consequence of this result for urban speed limits.
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.
- a) Calculate the length of the plane covered by the block and the normal force acting on it, $N = m \cdot g \cdot \cos \theta$.
- b) Calculate the friction force and its work done, $W_{roz} = -F_{roz} \cdot L$.
- c) Apply $W_{roz} = \Delta E_m$ to calculate the kinetic energy of the block when reaching the bottom and its speed.
- d) What percentage of the initial potential energy has been dissipated by friction?
Recognising when mechanical energy is conserved
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:
- a) A satellite in a circular orbit around the Earth.
- b) A car braking hard to a complete stop.
- c) An apple falling from a branch (ignore friction).
- d) A lift ascending at a constant speed.
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.
- a) Calculate its initial potential energy relative to the lowest point.
- b) Apply the conservation of energy and calculate the speed with which it passes through the lowest point.
- c) What height will it reach on the other side of the swing? Justify your answer without performing further calculations.
Back to the beginning: in the valley of the roller coaster all the initial potential energy has been transformed into kinetic energy; since $m \cdot g \cdot h = \frac{1}{2} m v^2$, the final speed is $v = \sqrt{2gh}$, a value that depends only on the height of the peak and not on the mass of the car or the shape of the track.