Theoretical Analysis of Energy and Work

1. Work and Kinetic Energy

Work ($W$) is the energy transferred to or from an object by a force acting on it that causes a displacement. It is a scalar quantity defined by the line integral of the force $\vec{F}$ along the path $C$ taken by the object from point $A$ to point $B$.

1.1. Formal Definition of Work

The Work done by a force $\vec{F}$ moving a particle along a path $C$ is:

For a constant force $\vec{F}$ and a straight displacement $\vec{d}$, this simplifies to the scalar product: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos\theta$, where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$.

1.2. Kinetic Energy (K)

Kinetic Energy ($K$) is the energy possessed by an object due to its motion. For a non-relativistic particle of mass $m$ and speed $v$, it is defined as:

In terms of linear momentum $\vec{p} = m\vec{v}$, kinetic energy can be expressed as $K = \frac{p^2}{2m}$.

1.3. The Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object by all external forces is equal to the change in its kinetic energy.

2. Potential Energy and Conservative Forces

2.1. Conservative Forces

A force $\vec{F}$ is defined as conservative if the work it does on a particle moving between two points is independent of the path taken. Mathematically, this is equivalent to stating that the work done around any closed loop is zero:

Alternatively, the force can be expressed as the negative gradient of a scalar potential energy function $U(\vec{r})$:

2.2. Gravitational Potential Energy ($U_g$)

Near the Earth's surface, the force of gravity is approximately constant $\vec{F}_g = -mg\hat{j}$. The potential energy associated with this force is proportional to height $y$ relative to a chosen reference level ($U=0$ at $y=0$):

For two general masses $M$ and $m$ separated by distance $r$, the gravitational potential energy is defined more accurately as:

2.3. Elastic Potential Energy ($U_s$)

For an ideal spring that obeys Hooke's Law ($\vec{F}_s = -k\vec{x}$), the stored potential energy due to compression or extension $x$ from equilibrium is:

3. Conservation of Mechanical Energy and Power

3.1. The Conservation of Mechanical Energy

Mechanical Energy ($E$) is the sum of Kinetic Energy ($K$) and Potential Energy ($U$): $E = K + U$. If only conservative forces perform work on a system, the mechanical energy is conserved.

If non-conservative forces ($F_{nc}$) are present (like friction or air resistance), the change in mechanical energy is equal to the work done by those forces:

3.2. Power (P)

Power is the rate at which work is done or the rate at which energy is transferred. Instantaneous power is the derivative of work with respect to time:

Using the definition of work, $dW = \vec{F} \cdot d\vec{r}$, and the relation $d\vec{r}/dt = \vec{v}$, instantaneous power can be expressed in terms of force and velocity:

4. Example: Path Independence of Gravitational Work

Let's demonstrate that the work done by the gravitational force $\vec{F}_g = -mg\hat{j}$ moving a particle from $A(x_1, y_1)$ to $B(x_2, y_2)$ is independent of the path.

4.1. Calculation via Direct Path (Straight Line)

The displacement vector is $d\vec{r} = dx\hat{i} + dy\hat{j}$. The work integral is:

The result depends only on the change in $y$ coordinates:

4.2. Calculation via a Step Path (Horizontal then Vertical)

Consider path $C_1$ (horizontal from $A(x_1, y_1)$ to $P(x_2, y_1)$) and $C_2$ (vertical from $P(x_2, y_1)$ to $B(x_2, y_2)$). $W = W_1 + W_2$.

The total work is $W = W_1 + W_2 = 0 + [-mg(y_2 - y_1)]$, which is identical to Eq. 4.1. This mathematically confirms the path independence of the gravitational force, demonstrating its conservative nature.

5. Practice Problems

The following problems explore applications of work, energy, and power principles across various dynamic systems.

5.1. Intermediate Problems (5)

1. A particle is subject to a variable force $\vec{F}(x) = (3x^2 - 2x) \hat{i}$. Calculate the total work done by this force as the particle moves from $x=1 \text{ m}$ to $x=3 \text{ m}$.
2. A block of mass $m=2.0 \text{ kg}$ is released from rest at a height $h=5.0 \text{ m}$. Use the conservation of mechanical energy to determine its speed $v$ just before it hits the ground, assuming negligible air resistance. (Use $g = 9.8 \text{ m/s}^2$).
3. A spring with constant $k=400 \text{ N/m}$ is compressed by $10 \text{ cm}$ ($\Delta x = 0.10 \text{ m}$). Calculate the maximum speed achieved by a $0.5 \text{ kg}$ mass attached to the end of the spring when it is released from rest on a smooth, horizontal surface.
4. A race car of mass $M$ accelerates uniformly from rest. If the engine supplies a constant power output $P_0$, find the speed $v$ of the car as a function of time $t$.
5. A force field is given by $\vec{F}(x, y) = 2xy \hat{i} + x^2 \hat{j}$. Show, using the curl condition, that this force field is conservative and derive its potential energy function $U(x, y)$.

5.2. Advanced Problems (3)

6. A block of mass $m$ slides down a rough incline of angle $\theta$. The coefficient of kinetic friction is $\mu_k$. The block starts from rest and slides a distance $L$. Using the relationship $W_{nc} = \Delta E$, derive the final velocity $v_f$ of the block in terms of $m, g, \theta, \mu_k$, and $L$.
7. A satellite of mass $m$ is orbiting a planet of mass $M$. Its speed at the farthest point (apoapsis), $r_a$, is $v_a$. Using the conservation of mechanical energy and angular momentum, find the speed $v_p$ at the closest point (periapsis), $r_p$.
8. A particle moves under the influence of a non-linear force $F(x) = c x^3$, where $c$ is a positive constant. If the particle is released from rest at $x_0$, find its velocity $v(x)$ as a function of position $x$.

5.3. Irodov-like Problems (2)

9. A flexible rope of mass $M$ and length $L$ hangs over the edge of a smooth table, with length $L/4$ initially hanging. Calculate the work done by gravity as the rope slides completely off the table.
10. A small body of mass $m$ is constrained to move along a track shaped as the curve $y = ax^2$ (where $a > 0$). Gravity acts vertically downward. Find the work done by the gravitational force when the body moves from $x_1$ to $x_2$.

6. Solutions to Practice Problems

Detailed, step-by-step solutions are provided below. Note the careful application of integrals and energy principles in each case.

6.1. Intermediate Solutions

Problem 1: Work Done by Variable Force

The work done is calculated directly from the line integral, since the force is parallel to the displacement ($d\vec{r} = dx \hat{i}$):

Problem 2: Gravitational Energy Conservation

Applying the conservation of mechanical energy ($E_i = E_f$), where $U_f = 0$ is set at the ground ($y=0$), and $K_i = 0$ since it starts from rest:

Problem 3: Elastic Energy Conservation

We equate the initial elastic potential energy ($U_s$) stored in the compressed spring to the final kinetic energy ($K$) of the mass, since $U_g$ is constant and $K_i=0$:

Problem 4: Constant Power Output

Power is defined as the rate of change of work, which is equal to the rate of change of kinetic energy. We use $P = dK/dt$:

Separating variables and integrating from $v=0$ to $v$ and $t=0$ to $t$:

Problem 5: Conservative Force and Potential

Conservatism Check (Curl Condition): $\nabla \times \vec{F} = \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{k}$.

Since the curl is zero, the force is conservative.

Potential Function: $\vec{F} = -\nabla U \implies F_x = -\partial U / \partial x$ and $F_y = -\partial U / \partial y$.

Equating this to the given $F_y = x^2$, we find $f'(y) = 0$, so $f(y) = C$ (a constant).

6.2. Advanced Solutions

Problem 6: Non-Conservative Work on an Incline

The non-conservative work done is by kinetic friction, $W_{nc} = W_{friction}$. The friction force is $f_k = \mu_k N = \mu_k (mg \cos\theta)$. The work is negative because it opposes displacement ($L$):

We use $W_{nc} = \Delta E = E_f - E_i$. Let $U=0$ at the bottom of the incline. The height change is $\Delta y = -L \sin\theta$.

Substituting into $W_{nc} = E_f - E_i$:

Problem 7: Satellite Motion (Energy and Angular Momentum)

Gravitational force is central, so both mechanical energy ($E$) and angular momentum ($L$) are conserved.

Energy Conservation ($E_a = E_p$):

Angular Momentum Conservation ($L_a = L_p$):

While the energy equation could be solved for $v_p$, the angular momentum conservation provides a direct and simpler proportional relationship between the velocities at apoapsis ($r_a$) and periapsis ($r_p$), as they are defined by their distance from the center of force, and the angular momentum is always perpendicular to the velocity vector at these two points:

Problem 8: Velocity from Non-Linear Force

We use the Work-Energy theorem ($W_{net} = \Delta K$). Since $\vec{F}$ is conservative (it is a function of position only), $W_{net} = -\Delta U$.

The potential energy function $U(x)$ is found by integration from a reference point $x_{ref}=0$ where $U=0$:

The change in energy from $x_0$ (rest, $v_0=0$) to $x$ is:

6.3. Irodov-like Solutions

Problem 9: Work Done by Gravity on a Sliding Rope

The work done by gravity is $W_g = -\Delta U_g$. The change in potential energy is calculated by considering the change in the center of mass (CM) of the hanging portion of the rope.

Initial CM of hanging part ($L/4$): $y_i = -L/8$. $U_i = m_i g y_i = \frac{M}{4} g \left(-\frac{L}{8}\right) = -\frac{M g L}{32}$.

Final CM of hanging part ($L$): $y_f = -L/2$. $U_f = M g y_f = -\frac{M g L}{2}$.

Work done:

Problem 10: Work Done on a Parabolic Track

Since gravity is a conservative force, the work done only depends on the change in gravitational potential energy, $W_g = -\Delta U_g = U_1 - U_2$.

Initial position: $y_1 = ax_1^2$. Final position: $y_2 = ax_2^2$.