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Additional: Work-Energy Theorem and Power



Work-Energy Theorem ($ W_{net} = \Delta KE $)

The Work-Energy Theorem is a fundamental principle in physics that directly links the concepts of work and kinetic energy. It provides a powerful tool for analyzing the motion of objects, stating how the total work done on an object affects its speed.


Statement of the Work-Energy Theorem

The theorem states that the net work done by all forces acting on an object is equal to the change in its kinetic energy.

Mathematically, this is expressed as:

$ W_{net} = \Delta KE $

where:

So, $ W_{net} = \frac{1}{2} m v_{final}^2 - \frac{1}{2} m v_{initial}^2 $


Implications of the Work-Energy Theorem

The Work-Energy Theorem applies regardless of whether the forces are constant or variable, or whether the path is straight or curved. However, the derivation is simpler for a constant force along a straight line.


Derivation of the Work-Energy Theorem (for Constant Net Force in 1D)

Consider an object of mass $m$ moving along a straight line (say, the x-axis) under the action of a constant net force $F_{net}$. Let the object's initial velocity be $v_i$ and final velocity be $v_f$ after undergoing a displacement $\Delta x$.

According to Newton's Second Law, the constant net force produces a constant acceleration $a$:

$ F_{net} = ma $

The work done by the constant net force over the displacement $\Delta x$ (assuming force and displacement are in the same direction, $\theta = 0^\circ$) is:

$ W_{net} = F_{net} \Delta x = (ma) \Delta x $

From the kinematic equation relating initial velocity ($v_i$), final velocity ($v_f$), constant acceleration ($a$), and displacement ($\Delta x$):

$ v_f^2 = v_i^2 + 2a \Delta x $

We can rearrange this equation to solve for $a \Delta x$:

$ v_f^2 - v_i^2 = 2a \Delta x $

$ a \Delta x = \frac{v_f^2 - v_i^2}{2} $

Now, substitute this expression for $a \Delta x$ back into the work done equation:

$ W_{net} = m (a \Delta x) = m \left(\frac{v_f^2 - v_i^2}{2}\right) $

$ W_{net} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 $

Recognizing the terms as the final and initial kinetic energies, we get:

$ W_{net} = KE_{final} - KE_{initial} = \Delta KE $

This derivation confirms the Work-Energy Theorem for the case of a constant net force acting in one dimension. The theorem is more general and holds true for variable forces and in three dimensions, though the derivation requires calculus.

Example 1. A car of mass 1200 kg accelerates from rest to a speed of 20 m/s. Calculate the net work done on the car.

Answer:

Mass of the car, $m = 1200$ kg.

Initial velocity, $v_{initial} = 0$ m/s (since it starts from rest).

Final velocity, $v_{final} = 20$ m/s.

Initial kinetic energy, $KE_{initial} = \frac{1}{2} m v_{initial}^2 = \frac{1}{2} \times 1200 \text{ kg} \times (0 \text{ m/s})^2 = 0$ J.

Final kinetic energy, $KE_{final} = \frac{1}{2} m v_{final}^2 = \frac{1}{2} \times 1200 \text{ kg} \times (20 \text{ m/s})^2 $

$ KE_{final} = 600 \times 400 $ J

$ KE_{final} = 240000 $ J

$ KE_{final} = 240 $ kJ

According to the Work-Energy Theorem, the net work done is the change in kinetic energy:

$ W_{net} = \Delta KE = KE_{final} - KE_{initial} $

$ W_{net} = 240000 \text{ J} - 0 \text{ J} = 240000 $ J

The net work done on the car is 240,000 Joules or 240 kJ.



Conservative and Non-Conservative Forces

Forces in physics can be broadly classified into two categories based on how the work they do depends on the path taken by the object:


Conservative Forces

A force is called conservative if the work done by the force in moving an object from one point to another depends only on the initial and final positions of the object, and is independent of the path taken between these points.

Equivalently, a force is conservative if the work done by the force in moving an object along any closed path (starting and ending at the same point) is zero.

For a conservative force, it is possible to define a corresponding potential energy. The work done by a conservative force is equal to the negative of the change in the associated potential energy:

$ W_c = - \Delta PE = -(PE_{final} - PE_{initial}) = PE_{initial} - PE_{final} $

This means that as a conservative force does positive work (e.g., gravity pulling an object down), the potential energy decreases. As work is done against a conservative force (e.g., lifting an object up against gravity), potential energy increases.

Examples of Conservative Forces:


Non-Conservative Forces

A force is called non-conservative if the work done by the force in moving an object from one point to another depends on the path taken between these points.

Equivalently, a force is non-conservative if the work done by the force in moving an object along a closed path is not zero.

Non-conservative forces typically dissipate mechanical energy from a system, often converting it into heat or sound. A potential energy function cannot be defined for a non-conservative force in the same way it can for a conservative force.

Examples of Non-Conservative Forces:


Distinction and Conservation of Energy

The distinction between conservative and non-conservative forces is crucial for the concept of conservation of mechanical energy. The principle of conservation of mechanical energy ($KE + PE = \text{constant}$) holds true only when the net work done by all non-conservative forces is zero.

If non-conservative forces do work, the total mechanical energy of the system changes. The work done by non-conservative forces ($W_{nc}$) is related to the change in both kinetic and potential energy:

$ W_{nc} = \Delta (KE + PE) = \Delta E_{mechanical} $

where $E_{mechanical} = KE + PE$. This equation essentially says that non-conservative forces are responsible for the change in the total mechanical energy of a system.

However, the total energy of an isolated system (including all forms, like thermal energy generated by friction) is always conserved. Non-conservative forces simply transform mechanical energy into other forms (like heat), rather than destroying it.

Example 1. A block is pushed across a rough horizontal surface. Identify the forces doing work and classify them as conservative or non-conservative. How does the mechanical energy of the block change?

Answer:

Forces acting on the block are:

  • Applied pushing force (let's assume horizontal).
  • Force of kinetic friction (horizontal, opposite to motion).
  • Gravitational force (vertical downwards).
  • Normal force from the surface (vertical upwards).

Assuming the block is pushed horizontally on a flat surface, the displacement is horizontal.

Let's analyse the work done by each force over a horizontal displacement $\Delta x$:

  • Applied pushing force: Does positive work if in the direction of motion, negative work if opposing. This is typically considered a non-conservative force as the energy input depends on the action of the external agent, not just the start/end points.
  • Force of kinetic friction: Acts opposite to the motion, so it does negative work ($W_f = -f_k \Delta x$). Friction depends on the path length travelled. Work done over a closed path is not zero. Therefore, friction is a non-conservative force.
  • Gravitational force: Acts vertically downwards. Since the displacement is horizontal ($\theta = 90^\circ$), the work done by gravity is zero ($W_g = mg \Delta x \cos 90^\circ = 0$). Gravity is a conservative force.
  • Normal force: Acts vertically upwards. Since the displacement is horizontal ($\theta = 90^\circ$), the work done by the normal force is zero ($W_N = N \Delta x \cos 90^\circ = 0$). The normal force, as a contact force supporting against motion perpendicular to the surface, doesn't do work in this scenario, and it's generally treated as non-conservative or non-working depending on the context (if it causes displacement, e.g., on an inclined plane, its work would depend on the normal force's interaction with the object's motion). In simple cases where it's perpendicular to displacement, it does no work.

The net work done on the block is $W_{net} = W_{applied} + W_{friction} + W_{gravity} + W_{normal} = W_{applied} + W_{friction} + 0 + 0 = W_{applied} + W_{friction}$.

According to the Work-Energy Theorem, $W_{net} = \Delta KE$. So, $W_{applied} + W_{friction} = \Delta KE$.

The total mechanical energy of the block is $E_{mechanical} = KE + PE_g$. Since the height is not changing on a horizontal surface, the gravitational potential energy $PE_g$ remains constant. Thus, $\Delta PE_g = 0$.

The change in mechanical energy is $\Delta E_{mechanical} = \Delta KE + \Delta PE_g = \Delta KE + 0 = \Delta KE$.

Comparing this with $W_{net} = \Delta KE$, we have $W_{applied} + W_{friction} = \Delta E_{mechanical}$.

If we separate conservative and non-conservative work: $W_c = W_{gravity} + W_{normal} = 0$. $W_{nc} = W_{applied} + W_{friction}$.

The total work done is $W_{net} = W_c + W_{nc} = 0 + W_{nc} = W_{nc}$.

So, $W_{nc} = \Delta KE$.

Also, we know $W_c = -\Delta PE_g$. Since $W_c = 0$, $\Delta PE_g = 0$.

The total work done is also $W_{net} = \Delta KE$. $W_{nc} = \Delta KE$.

We can also relate non-conservative work directly to the change in mechanical energy:

$ W_{nc} = \Delta (KE + PE_g) = \Delta KE + \Delta PE_g $

In this horizontal motion case, $\Delta PE_g = 0$, so $W_{nc} = \Delta KE$. Since $W_{friction}$ is negative, $W_{applied} + W_{friction} = \Delta KE$. If the applied work is just balancing friction ($W_{applied} = -W_{friction} > 0$), $\Delta KE = 0$ and the object moves at constant velocity, but mechanical energy is being converted to heat by friction ($|W_{friction}|$ amount) and replaced by the applied force doing work ($W_{applied}$ amount). The net non-conservative work ($W_{applied} + W_{friction}$) determines the change in KE.

In summary, friction is a non-conservative force. It does negative work on the block, leading to a decrease in the mechanical energy ($KE+PE_g$) of the block if there is no other energy input (like the applied force). The presence of the non-conservative force of friction means that the mechanical energy of the block itself is not conserved; some of it is converted into thermal energy.



Power as Force times Velocity ($ P = \vec{F} \cdot \vec{v} $)

We previously defined power as the rate at which work is done ($P = dW/dt$) or energy is transferred. A useful alternative expression for instantaneous power relates it directly to the force acting on an object and its instantaneous velocity.


Derivation of $P = \vec{F} \cdot \vec{v}$

Consider a force $\vec{F}$ acting on an object. Let the object undergo a small displacement $d\vec{r}$ in a small time interval $dt$.

The infinitesimal work done by the force during this small displacement is:

$ dW = \vec{F} \cdot d\vec{r} $

The instantaneous power $P$ is the rate of doing work, which is the work done $dW$ divided by the time interval $dt$:

$ P = \frac{dW}{dt} $

Substitute the expression for $dW$ into the equation for power:

$ P = \frac{\vec{F} \cdot d\vec{r}}{dt} $

Using the properties of the dot product and the definition of instantaneous velocity $\vec{v} = \frac{d\vec{r}}{dt}$, we can write:

$ P = \vec{F} \cdot \left(\frac{d\vec{r}}{dt}\right) $

$ P = \vec{F} \cdot \vec{v} $

This formula, $P = \vec{F} \cdot \vec{v}$, gives the instantaneous power delivered by the force $\vec{F}$ to the object that has instantaneous velocity $\vec{v}$.


Physical Interpretation

Since the dot product can be written as $P = F v \cos\theta$, where $\theta$ is the angle between $\vec{F}$ and $\vec{v}$, this formula shows:

This formula is particularly useful when dealing with situations where force and velocity are known vectors, or when power output is related to the speed of a system (e.g., power of an engine needed to maintain a certain speed against resistive forces).

Example 2. A motor is pulling a cart with a constant force of 500 N at a constant velocity of 2 m/s in the same direction as the force. Calculate the power delivered by the motor.

Answer:

Force applied by the motor, $F = 500$ N.

Velocity of the cart, $v = 2$ m/s.

The force and velocity are in the same direction, so the angle between them is $\theta = 0^\circ$.

Using the formula $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$:

$ P = 500 \text{ N} \times 2 \text{ m/s} \times \cos(0^\circ) $

$ P = 500 \times 2 \times 1 $ Watts

$ P = 1000 $ W

This can also be expressed as 1 kilowatt (kW).

The power delivered by the motor is 1000 Watts or 1 kW.

Example 3. A block of mass 5 kg is pulled up a smooth inclined plane at a constant speed of 0.5 m/s by a force parallel to the incline. The incline makes an angle of 30° with the horizontal. Calculate the power delivered by the pulling force. (Take $g = 9.8 \, \text{m/s}^2$).

Answer:

Mass of the block, $m = 5$ kg.

Constant speed, $v = 0.5$ m/s.

Angle of incline, $\alpha = 30^\circ$.

$g = 9.8 \, \text{m/s}^2$.

Since the block moves at a constant speed, its acceleration is zero. By Newton's Second Law, the net force along the incline must be zero.

Forces acting on the block along the incline are:

  • The pulling force $\vec{F}_P$ (up the incline).
  • The component of gravity parallel to the incline, $mg\sin\alpha$ (down the incline).

Let the pulling force be $F_P$. Since net force along the incline is zero:

$ F_P - mg\sin\alpha = 0 $

$ F_P = mg\sin\alpha $

$ F_P = 5 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(30^\circ) $

$ F_P = 5 \times 9.8 \times 0.5 $ N

$ F_P = 24.5 $ N

The pulling force is parallel to the velocity (up the incline), so the angle $\theta$ between $\vec{F}_P$ and $\vec{v}$ is $0^\circ$.

The power delivered by the pulling force is $P = \vec{F}_P \cdot \vec{v} = F_P v \cos\theta$

$ P = 24.5 \text{ N} \times 0.5 \text{ m/s} \times \cos(0^\circ) $

$ P = 24.5 \times 0.5 \times 1 $ Watts

$ P = 12.25 $ W

The power delivered by the pulling force is 12.25 Watts.