Particles Work Energy Principle
Starting from Euler’s first law written as:
\sum \mathbf F = M \mathbf A
and taking the scalar product with the velocity of the center of mass \mathbf V:
\sum \mathbf F \cdot \mathbf V = M \mathbf A \cdot \mathbf V = M \frac{\mathrm d \mathbf V}{\mathrm dt} \cdot \mathbf V = \frac{M}{2} \frac{\mathrm d (\mathbf V \cdot \mathbf V)}{\mathrm dt}
Integrating both sides from t_1 to t_2:
\begin{aligned} & \int_{t_1}^{t_2} \sum \mathbf F \cdot \mathbf V = \int_{t_1}^{t_2}\frac{M}{2} \frac{\mathrm d (\mathbf V \cdot \mathbf V)}{\mathrm dt} = \frac{M}{2} \left( |\mathbf V(t_2)|^2 - |\mathbf V(t_1)|^2 \right) \\ & \int_{t_1}^{t_2} \sum \mathbf F \cdot \frac{\mathrm d \mathbf R}{\mathrm dt} = \sum \mathbf F \cdot \left( \mathbf R(t_2) - \mathbf R(t_1) \right) = \frac{M}{2} \left( |\mathbf V(t_2)|^2 - |\mathbf V(t_1)|^2 \right) \end{aligned}
The left hand side is the sum of forces dotted with distances, so it is the difference of the work of the external forces on the body, and the right hand side it is a change in translational kinetic energy:
W = \Delta T
Work is a scalar quantity, meaning it has magnitude but no direction. Its value depends on the force applied and the displacement in the direction of the force, calculated as W = F \cdot d \cdot \cos \theta, where \theta is the angle between the force and displacement vectors.
The sign convention for work is as follows:
- positive work: When 0^\circ \leq \theta < 90^\circ, the force component and displacement are in the same direction, and energy is transferred to the object (e.g., lifting an object upward);
- negative work: When 90^\circ < \theta \leq 180^\circ, the force component opposes the displacement, taking energy from the object (e.g., friction slowing down an object);
- zero work: When \theta = 90^\circ, the force is perpendicular to displacement, and no energy is transferred (e.g., holding an object steady without movement).
Kinetic energy is also a scalar quantity, representing the energy of an object due to its motion. It is given by KE = \frac{1}{2}mv^2, where m is the mass and v is the velocity of the object.
For kinetic energy:
- Positive kinetic energy: Kinetic energy is always positive because it depends on the square of the velocity, v^2, making it independent of the direction of motion;
- zero kinetic energy: When an object is at rest (i.e., v = 0), its kinetic energy is zero because there is no motion.
Kinetic energy does not have a negative value, as it only measures the magnitude of motion energy, not direction.
Let’s study the work of some different forces. Force can be of two types: conservative and non-conservative.
A conservative force is a force for which the work done is independent of the path taken. The work done by a conservative force only depends on the initial and final positions. For conservative forces, there exists a potential energy function U such that the force can be derived from it:
\vec{F} = -\nabla U
Examples of conservative forces include gravitational force, electrostatic force, and spring force.
Key properties of conservative forces:
- Path Independence: The work done by a conservative force between two points is the same regardless of the path.
- Closed Loop: The work done by a conservative force in any closed loop is zero.
- Potential Energy: Conservative forces have an associated potential energy.
A non-conservative force is a force for which the work done depends on the path taken. This means that the work done around a closed loop is not zero and energy is generally dissipated, often as heat or friction.
Examples of non-conservative forces include friction, air resistance, and viscous forces.
Key properties of non-conservative forces:
- Path Dependence: The work done depends on the path taken.
- Energy Dissipation: Non-conservative forces dissipate energy, which is usually converted to other forms like thermal energy.
- No Potential Energy: There is no potential energy function for non-conservative forces, as the work done cannot be fully recovered.
Work of a spring
Let’s consider a linear spring.
The spring has a stiffness constant k and it is attached to a fixed wall on one end, and two positions, \text{Position}_1 and \text{Position}_2, representing different states along the spring’s extension. The spring’s natural length (unstretched length) is denoted by L_0, while L_1 and L_2 represent the lengths corresponding to \text{Position}_1 and \text{Position}_2, respectively. The displacements \delta_1 and \delta_2 are the extensions of the spring from its natural length, given by \delta_1 = L_1 - L_0 and \delta_2 = L_2 - L_0.
The spring exerts a conservative force that obeys Hooke’s Law:
F = -k \delta
where \delta = L - L_0 is the displacement from the natural length L_0. The work W done by the spring force when moving from position L_1 to L_2 can be calculated by integrating the force over the displacement:
W = \int_{L_1}^{L_2} F \, dL = - \int_{L_1}^{L_2} k (L - L_0) \, dL
Evaluating this integral, we get:
W = -k \int_{L_1}^{L_2} (L - L_0) \, dL = -\frac{k}{2} \left( (L_2 - L_0)^2 - (L_1 - L_0)^2 \right)
This expression represents the work done by the conservative spring force as the object moves from \text{Position}_1 (at L_1) to \text{Position}_2 (at L_2).
Since the spring force is conservative, we can define a potential energy function U associated with it. The potential energy U of a spring at a given displacement \delta = L - L_0 is:
U(L) = \frac{1}{2} k (L - L_0)^2
The work done by the spring force when moving from L_1 to L_2 can also be expressed as the negative change in potential energy:
W = U(L_1) - U(L_2)
Substituting the expressions for U(L_1) and U(L_2):
W = \frac{1}{2} k (L_1 - L_0)^2 - \frac{1}{2} k (L_2 - L_0)^2 = -\frac{k}{2} \left[ \delta(t_2)^2 - \delta(t_1)^2\right]
This result is consistent with our previous calculation for the work done, confirming that the work done by the spring force is equal to the decrease in the potential energy of the system as the spring moves from \text{Position}_1 to \text{Position}_2.
Work by gravity
For gravitational work, consider a mass m moving vertically from an initial position y_1 to a final position y_2 under the influence of gravity with acceleration g. The gravitational force F_g acting on the mass is given by:
F_g = -mg
The work W done by the gravitational force as the object moves from y_1 to y_2 is calculated by integrating this force over the vertical displacement:
W = \int_{y_1}^{y_2} F_g \, dy = - \int_{y_1}^{y_2} mg \, dy
Evaluating the integral, we get:
W = -mg \int_{y_1}^{y_2} dy = -mg (y_2 - y_1)
Thus, the work done by gravity when moving from height y_1 to y_2 is:
W = -mg (y_2 - y_1)
This expression represents the work done by the gravitational force as the object moves between two vertical positions y_1 and y_2.
Work by Coulomb friction
Let’s consider a block sliding on a surface, subject only to the weight \mathbf W = m \mathbf g.
The friction force \mathbf{f} acting on a block sliding on a surface with a normal force \mathbf{N} is given by:
\mathbf{f} = \mu \mathbf{N}
where \mu is the coefficient of friction. For a block moving horizontally over a distance d from \text{Position}_1 to \text{Position}_2, the work W_f done by the friction force is calculated as:
W_f = \int_{0}^{d} \mathbf{f} \cdot d\mathbf{s} = - \mu \mathbf N d
where d is the distance traveled and the negative sign reflects that the frictional force opposes the direction of motion.
If the block moves from \text{Position}_1 to \text{Position}_2 and then returns to the starting point, the work done by friction over each leg of the trip is calculated as:
W_{f, \, \text{round trip}} = W_f + W_f = - \mu \mathbf N d - \mu \mathbf N d = -2 \mu \mathbf N d
The total work is twice the work done in a single leg because friction always acts opposite to the direction of motion. Regardless of the direction, frictional work is path-dependent and always negative, dissipating energy as heat. For a round trip, the work done by friction is twice the work calculated for a single trip and not zero as it would be for a conservative force.
Work-energy problems
Let’s consider a block sliding on a surface, subject only to the weight \mathbf W = m \mathbf g, and use the work-energy principle to solve this sample problem.
We have a mass m = 12 \, \text{kg}, initial velocity v_0 = 1.9 \, \text{m/s}, distance traveled d = 2.5 \, \text{m}, a coefficient of kinetic friction \mu = 0.3, a spring constant k = 20 \, \text{kN/m} = 20000 \, \text{N/m}, and it slides on an inclined slope \frac{3}{4}.
We know that \tan(\theta) = \frac{3}{4}, so:
\begin{aligned} \sin(\theta) & = \frac{3}{5} \\ \cos(\theta) & = \frac{4}{5} \end{aligned}
The initial kinetic energy T_0 of the block is:
T_0 = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 12 \times (1.9)^2 = \frac{1}{2} \times 12 \times 3.61 = 21.66 \, \text{J}
The work done by friction W_f is:
W_f = -\mu \cdot m \cdot g \cdot d \cdot \cos(\theta) = -0.3 \times 12 \times 9.81 \times 2.5 \times \frac{4}{5} = -70.632 \, \text{J}
The work by to gravity W_g is the change in potential energy due to the height change is:
W_g = m \cdot g \cdot d \cdot \sin(\theta) = 12 \times 9.81 \times 2.5 \times \frac{3}{5} = 176.58 \, \text{J}
The work done by the spring is the unknown to compute, but the formula is known (negative because it oppose to the motion and we can set the reference of the spring compression to zero at rest (x_0 = 0):
W_s = -\frac{1}{2} \left (x_1^2 - x_0^2\right) = -\frac{1}{2} x_1^2
Using the work-energy principle, the work done by the external force is equal to the change in kinetic energy and the maximum force is with the spring at maximum compression (when v = 0):
\begin{aligned} & W = \Delta T \\ & W_g + W_f + W_s = \Delta T = T_1 - T_0 = -T_0 \end{aligned}
The only unknown is the work of the spring:
T_0 + W_g + W_f = -W_s = \frac{1}{2} k x_{\text{max}}^2
Substituting the values:
\begin{aligned} & 21.66 + 176.58 - 70.632 = \frac{1}{2} \times 20000 \times x_{\text{max}}^2 \\ & 127.608 = 10000 \times x_{\text{max}}^2 \\ & x_{\text{max}}^2 = \frac{127.608}{10000} \\ & x_{\text{max}} = \sqrt{0.0127608} \\ & x_{\text{max}} \approx 0.113 \, \text{m} \end{aligned}
The maximum force F_{\text{max}} in the spring is:
F_{\text{max}} = k \cdot x_{\text{max}} = 20000 \times 0.113 = 2260 \, \text{N}
The maximum force in the spring is approximately 2260 \, \text{N}.