Interactive Lecture
Work, Kinetic and Potential Energy, and Power
by Dr Arnaldo Delli Carri
The principle of work and energy is a fundamental concept in mechanics that relates the work done on a body to the change in its kinetic energy. This principle provides a powerful alternative to Newton's second law for solving dynamics problems.
Energy methods often simplify problems by eliminating the need to consider forces that do no work, and by avoiding the direct calculation of acceleration.
Work of a Force:
For a force \(\vec{F}\) acting through an infinitesimal displacement \(d\vec{s}\) from position \(s_1\) to position \(s_2\):
\(W = \displaystyle \int_{s_1}^{s_2}\vec{F} \cdot d \vec{s}\)
where \(\cdot\) is the scalar product between the force and displacement vectors, therefore the work done is a scalar quantity, and its SI unit is the Joule \([\text{J}]=[\text{Nm}]\).
Work of a Couple Moment:
For a constant couple moment \(M\) (in \(\text{N}\cdot\text{m}\)) acting through an angular displacement \(\theta\) (in radians):
\(W = M \theta\)
When a constant force acts on a particle that moves along a straight line, the work done is simply the product of the force component in the direction of motion and the distance travelled, or alternatively by the product of the force and the component of distance travelled in the same direction of the force.
\(W_F = \vec{F} \cdot \Delta\vec{s}=\big(F\cos\theta\big)\Delta s =F\big(\Delta s \cos\theta\big)\)
where \(\theta\) is the angle between the force vector and the displacement vector.
The gravitational force is a special case of a constant force (near Earth's surface):
For a mass \(m\) moving vertically through height \(h\):
\(W_{g} = -mgh\) (when moving upward)
\(W_{g} = +mgh\) (when moving downward)
Important: The work done by gravity depends only on the vertical displacement, not on the path taken. This makes gravity a conservative force.
The force on a spring is proportional to its stiffness (spring coefficient \(k\)) and its displacement (extension):
\(F_k=-ks\)
For a spring being extended from \(s_1\) to \(s_2\):
\(W_{k} = \displaystyle \int_{s_1}^{s_2}-ks \, ds=-\frac{1}{2}k\big( s_2^2-s_1^2\big) \)
Reaction forces \(F_R\) are always perpendicular to the direction of motion, therefore they do no work:
Since \(\theta = 90°\) between normal force and displacement:
\(W_{R} = F_R \cdot \Delta s \cdot \cos(90°) = 0\)
This is why energy methods are particularly useful: forces such as reactions that do no work can be left out of the energy balance entirely.
Friction forces always oppose motion and therefore do negative work:
\(W_{f} = -F_f \cdot \Delta s = -\mu_k N \cdot \Delta s\)
Conservative vs Non-Conservative Forces:
For conservative forces, we can define a potential energy function:
Gravitational Potential Energy: \(U_g = mgh\)
Elastic Potential Energy: \(U_e = \frac{1}{2}kx^2\)
The work done by a conservative force equals the negative change in potential energy:
\(W_{\text{cons}} = -\Delta U\)
By integrating Newton's second law with respect to displacement
\[\displaystyle \int \sum F \, ds=\displaystyle \int ma \, ds\]
\[\displaystyle \sum \int F \, ds=\displaystyle \int mv \, dv\]
\[\displaystyle \sum W = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\]
hence we define the quantity kinetic energy \(E_k=\frac{1}{2}mv^2\)
and we arrive at
The Work-Energy Theorem:
\[\sum W = \Delta E_k\]
The total work done on a particle equals the change in its kinetic energy.
Power is the rate at which work is done, or equivalently the rate at which energy is transferred:
Power:
\(P = \dfrac{dW}{dt}\)
For a force \(\vec F\) acting on a body moving with velocity \(\vec v\):
\(P = \vec F \cdot \vec v\)
The SI unit of power is the watt, \([\text{W}] = [\text{J/s}]\).
For rotating rigid bodies:
Rotational Kinetic Energy: \(E_{k,\text{rot}} = \frac{1}{2}I\omega^2\)
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
Total Energy for Rolling Motion:
\(E_{k,\text{tot}} = E_{k,\text{trans}} + E_{k,\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\)
starting from the previous form:
\[\sum W = \Delta E_k\]
we can split the sum of work into work done by conservative forces \(W_{\text{cons}}\) and work done by non-conservative forces \(W_{\text{nc}}\)
\[W_{\text{cons}}+W_{\text{nc}} = \Delta E_k\]
recalling the definition of potential energy for conservative forces \(W_{\text{cons}}=-\Delta U\)
\[-\Delta U+W_{\text{nc}} = \Delta E_k\]
finally, by noting that \(\Delta U=U_2-U_1\,\) and \(\,\Delta E_k=E_{k2}-E_{k1}\,\) we arrive at an alternative form of the Work-Energy theorem
\[E_{k1}+U_1+W_{\text{nc}} = E_{k2}+U_2\]
The final mechanical (kinetic and potential) energy of the system is equal to the initial mechanical energy of the system plus (minus) the contribution of non-conservative forces.
\[E_1+W_{\text{nc}} = E_2\]
corollary: in the absence of non-conservative forces (such as friction and drag), the total mechanical energy of the system is conserved \(E_1=E_2\).
Release the bead from different heights and observe how friction affects the maximum height reached. Toggle friction to see the difference between conservative and non-conservative systems.
Observation: Without friction, the bead returns to its initial height (energy is conserved). With friction, some energy is dissipated as heat, and the bead doesn't reach the initial height.
Watch how potential energy converts to kinetic energy as the block slides down. The energy bars show both absolute (Joules) and percentage values, including energy dissipated by friction.
Compare how different shaped objects roll down an incline. Each object has the same mass and radius, but a different moment of inertia, leading to a different distribution of rotational and translational kinetic energy.
Objects with lower moments of inertia accelerate faster down the incline because less energy goes into rotation and more stays translational. Solid spheres beat hollow ones every time — the acceleration doesn't depend on mass or radius, only on how the mass is distributed.
Problem-Solving Strategy:
| Concept | Equation | Description |
|---|---|---|
| Work of a Force | \(W = \vec{F} \cdot \vec{s}\) | Scalar product of force and displacement |
| Kinetic Energy (Translation) | \(E_k = \frac{1}{2}mv^2\) | Energy due to linear motion |
| Kinetic Energy (Rotation) | \(E_{k,\text{rot}} = \frac{1}{2}I\omega^2\) | Energy due to rotational motion |
| Gravitational PE | \(U_g = mgh\) | Potential energy due to height |
| Elastic PE | \(U_e = \frac{1}{2}kx^2\) | Energy stored in springs |
| Work-Energy Theorem | \(\sum W = \Delta E_k\) | Fundamental principle |
| Power | \(P = \vec F \cdot \vec v\) | Rate of doing work |
| Conservation of Energy | \(E_1 + W_{\text{nc}} = E_2\) | Including non-conservative work |
A 5 kg block starts from rest at the top of a 30° incline that is 2 metres long. If the coefficient of kinetic friction is 0.2, find the speed of the block at the bottom.
Show Solution
Given: \(m = 5\) kg, \(\theta = 30°\), \(s = 2\) m, \(\mu_k = 0.2\), \(v_1 = 0\)
Height change: \(h = s \sin\theta = 2 \sin 30° = 1\) m
Normal force: \(N = mg\cos\theta = 5 \times 9.81 \times \cos 30° = 42.5\) N
Work by friction: \(W_f = -\mu_k N s = -0.2 \times 42.5 \times 2 = -17\) J
Using work-energy theorem:
\(mgh + W_f = \frac{1}{2}mv_2^2\)
\(5 \times 9.81 \times 1 - 17 = \frac{1}{2} \times 5 \times v_2^2\)
\(49.05 - 17 = 2.5v_2^2\)
\(v_2 = \sqrt{\frac{32.05}{2.5}} = 3.58\) m/s
Answer: \(v_2 = 3.58\) m/s
A solid cylinder of mass 10 kg and radius 0.2 m rolls without slipping down a 45° incline. If it starts from rest and rolls 3 metres along the incline, find its final linear velocity.
Show Solution
Given: \(m = 10\) kg, \(R = 0.2\) m, \(\theta = 45°\), \(s = 3\) m
For a solid cylinder: \(I = \frac{1}{2}mR^2\)
Height change: \(h = s\sin\theta = 3\sin 45° = 2.12\) m
For rolling without slipping: \(v = R\omega\)
Energy equation:
\(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\)
\(mgh = \frac{1}{2}mv^2 + \frac{1}{2} \times \frac{1}{2}mR^2 \times \frac{v^2}{R^2}\)
\(gh = \frac{1}{2}v^2 + \frac{1}{4}v^2 = \frac{3}{4}v^2\)
\(v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4 \times 9.81 \times 2.12}{3}} = 5.27\) m/s
Answer: \(v = 5.27\) m/s
A 2 kg mass is attached to a horizontal spring with stiffness k = 400 N/m. The mass is pulled 0.15 m from equilibrium and released on a surface with coefficient of kinetic friction μ = 0.2. How far does the mass travel before coming to rest?
Show Solution
Given: \(m = 2\) kg, \(k = 400\) N/m, \(x_0 = 0.15\) m, \(\mu_k = 0.2\)
Initial energy (all elastic PE): \(E_0 = \frac{1}{2}kx_0^2 = \frac{1}{2} \times 400 \times 0.15^2 = 4.5\) J
Friction force: \(F_f = \mu_k mg = 0.2 \times 2 \times 9.81 = 3.92\) N
Let total distance travelled = d. Work done by friction: \(W_f = -F_f \times d = -3.92d\)
Using work-energy theorem: \(E_0 + W_f = 0\) (final KE and PE both zero)
\(4.5 - 3.92d = 0\)
\(d = \frac{4.5}{3.92} = 1.15\) m
Note: The mass will oscillate with decreasing amplitude. The spring will compress and extend multiple times, with the total distance travelled being 1.15 m before stopping.
Answer: \(d = 1.15\) m