INTRODUCTION
Everything covered in this lesson, in one place - useful for revision or printing.
Mechanics Energy Potential, kinetic and mechanical energy
Energy is what lets an object do work: a book on a high shelf, a runner, a swinging pendulum and a roller coaster all have energy.
In this lesson you will learn to:
Energy of position Measured from a reference point Symbol: \(E_p\) (or \(U\))
Gravitational potential energy is the energy an object has because of its position in the gravitational field relative to some reference point.
The higher an object is above the reference point, the more gravitational potential energy it has. You choose the reference point (often the floor or the ground) and call its height \(h = 0\).
\[E_p = mgh\] m in kg, g = 9,8 m·s -2 , h in m, E p in J
Calculate gravitational potential energy with \(E_p = mgh\):
A 2 kg book on a shelf 1,5 m above the floor. The floor is the reference level. 2 kg h = 1,5 m Floor: reference level (h = 0) Drawn to scale Height is measured from the chosen reference level, here the floor.
A 2 kg book is on a shelf 1,5 m above the floor. Calculate its gravitational potential energy relative to the floor.
Quick check: Calculate \(E_p\) of the book relative to the floor, in J.
\(E_p\): energy of position \(E_p = mgh\) Book: \(E_p = 29{,}4\ \text{J}\)
Summary. Gravitational potential energy depends on mass and on height above a chosen reference point. Always state the reference point, and always use SI units: kg, m and J.
Energy of motion Moving objects only Symbol: \(E_k\) (or \(K\))
Kinetic energy is the energy an object has as a result of its motion.
An object at rest has zero kinetic energy. The faster it moves, and the more mass it has, the more kinetic energy it has.
Quick check: A ball is held still at the top of a building. What is its kinetic energy?
\[E_k = \tfrac{1}{2}mv^2\] m in kg, v in m·s -1 , E k in J
Calculate kinetic energy with \(E_k = \tfrac{1}{2}mv^2\):
Worked example 60 kg runner moving at 5 m·s -1
A 60 kg runner is moving at 5 m·s-1. Calculate the runner's kinetic energy.
Quick check: Calculate \(E_k\) of the runner, in J.
\(E_k\): energy of motion \(E_k = \tfrac{1}{2}mv^2\) Runner: \(E_k = 750\ \text{J}\)
Summary. Kinetic energy depends on mass and on the square of the speed. An object at rest has zero kinetic energy.
\[E_M = E_k + E_p\] also written E M = K + U
Mechanical energy is the sum of the gravitational potential energy and the kinetic energy of an object.
Calculate it with \(E_M = E_k + E_p\), or in the other notation \(E_M = K + U\).
Worked example 0,5 kg ball 4 m above the ground, moving at 6 m·s -1
A 0,5 kg ball is 4 m above the ground and moving at 6 m·s-1. Calculate its kinetic energy, its gravitational potential energy (relative to the ground) and its mechanical energy.
Quick check: First, calculate \(E_k\) of the ball, in J.
Quick check: Next, calculate \(E_p\) of the ball, in J.
\[E_M = 9 + 19{,}6 = 28{,}6\ \text{J}\]
Mechanical energy is the sum of the two energies you just found.
Quick check: Calculate \(E_M\) of the ball, in J.
\(E_M = E_k + E_p\) Ball: \(E_k = 9\ \text{J}\), \(E_p = 19{,}6\ \text{J}\) Ball: \(E_M = 28{,}6\ \text{J}\)
Summary. To find mechanical energy, calculate \(E_k\) and \(E_p\) separately, then add them.
The total energy in an isolated system remains constant.
Law of conservation of energy: the total energy in an isolated system remains constant.
Energy is not created or destroyed; it changes from one form to another. When friction acts, some mechanical energy changes into heat and sound, but the total energy is still the same.
\[(E_p + E_k)_{\text{top}} = (E_p + E_k)_{\text{bottom}}\]
Principle of conservation of mechanical energy: in the absence of air resistance or any external forces, the mechanical energy of an object is constant.
So potential energy and kinetic energy can swap, but their sum \(E_M\) stays the same. You can write \(E_{M,\text{top}} = E_{M,\text{bottom}}\).
Start: \(mgh_1 + \tfrac{1}{2}mv_1^2\) equals End: \(mgh_2 + \tfrac{1}{2}mv_2^2\)
The method for every conservation problem:
Quick check: When is mechanical energy conserved?
Total energy of an isolated system is constant No friction / air resistance: \(E_M\) is constant \(E_p\) and \(E_k\) swap; their sum stays the same
Summary. Know both statements word for word: the law of conservation of energy (all energy, isolated system) and the principle of conservation of mechanical energy (\(E_M\) constant when there is no air resistance or external force).
\[mgh = \tfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{2gh}\]
A ball is dropped from rest from 5 m. Calculate its speed just before it hits the ground.
Assume no friction / air resistance, so mechanical energy is conserved. Take the ground as h = 0. At the top it has only \(E_p\) (it starts from rest); just before the ground it has only \(E_k\). The mass appears on both sides, so it cancels.
Quick check: Calculate the speed just before it hits the ground, in m·s-1 (2 decimals).
\[\tfrac{1}{2}mv^2 = mgh \;\Rightarrow\; h = \dfrac{v^2}{2g}\]
A ball is thrown vertically upwards at 12 m·s-1. Calculate its maximum height.
Assume no friction / air resistance, so mechanical energy is conserved. At the launch point (h = 0) it has only \(E_k\). At maximum height it stops for an instant, so it has only \(E_p\).
Quick check: Calculate the maximum height, in m (2 decimals).
Dropped from 5 m: \(v = 9{,}90\ \text{m·s}^{-1}\) Thrown up at 12 m·s -1 : \(h = 7{,}35\ \text{m}\) The mass cancels in both
Summary. For vertical motion with no air resistance, \(E_p\) at the top equals \(E_k\) at the bottom. At maximum height the speed is zero; when dropped from rest, the starting speed is zero.
A pendulum bob released from rest 0,2 m above its lowest point. 0,2 m Released from rest Lowest point (h = 0) Drawn to scale The bob swings down through a height of 0,2 m to its lowest point.
A pendulum bob is released from rest 0,2 m above its lowest point. Calculate its speed at the lowest point.
Assume no friction / air resistance, so mechanical energy is conserved. Take the lowest point as h = 0: all the \(E_p\) at the release point becomes \(E_k\) at the bottom.
Quick check: Calculate the speed at the lowest point, in m·s-1 (2 decimals).
Roller-coaster track: the cart starts from rest at A on a 20 m hill and passes point B, 8 m above the ground. 20 m A (rest) B 8 m Ground (h = 0) Drawn to scale From A (20 m) to B (8 m) the cart drops 20 - 8 = 12 m.
A 500 kg roller-coaster cart starts from rest at the top of a 20 m hill (A). Calculate its speed at B, 8 m above the ground.
Assume no friction / air resistance, so mechanical energy is conserved. Take the ground as h = 0. Write \(E_{M,A} = E_{M,B}\):
\(mgh_A + 0 = mgh_B + \tfrac{1}{2}mv_B^2\)
The mass cancels (every term has m), so the 500 kg is not needed: \(v_B^2 = 2g(h_A - h_B)\), and the drop is 20 - 8 = 12 m.
Quick check: Calculate the speed at B, in m·s-1 (2 decimals).
A 3 kg block at the top of a frictionless slope 1,2 m high. 1,2 m 3 kg, starts from rest No friction Bottom of slope (h = 0) Drawn to scale The block slides from a height of 1,2 m to the bottom of the slope.
A 3 kg block slides from rest down a frictionless slope from a height of 1,2 m. Calculate its speed at the bottom.
Assume no friction / air resistance, so mechanical energy is conserved. Only the vertical height matters, not the length of the slope. Take the bottom as h = 0; the mass cancels again.
Quick check: Calculate the speed at the bottom, in m·s-1 (2 decimals).
Pendulum (0,2 m): \(v = 1{,}98\ \text{m·s}^{-1}\) Coaster (12 m drop): \(v = 15{,}34\ \text{m·s}^{-1}\) Slope (1,2 m): \(v = 4{,}85\ \text{m·s}^{-1}\)
Summary. Pendulums, roller coasters and slopes all use the same idea: with no friction / air resistance, \(E_M\) at the start equals \(E_M\) at the end. Use vertical heights only; the mass cancels.
\(E_p = mgh\) \(E_k = \tfrac{1}{2}mv^2\) \(E_M = E_k + E_p\) No friction / air resistance: \(E_M\) is constant
Energy: full summary