Work done by constant and variable forces, kinetic and potential energies

Work Done by Constant Force

In physics, work is a fundamental concept that describes the transfer of energy. When a force acts on an object and causes it to move over a distance, work is done. For a constant force, the work done is defined as the product of the magnitude of the force and the distance moved in the direction of the force.

Mathematically, if a constant force $\vec{F}$ acts on an object, and the object undergoes a displacement $\vec{d}$, the work done $W$ is given by the dot product of the force and displacement vectors:

$W = \vec{F} \cdot \vec{d}$

This can also be written as:

$W = |\vec{F}| |\vec{d}| \cos \theta$

Where:

• $W$ is the work done.
• $|\vec{F}|$ is the magnitude of the constant force.
• $|\vec{d}|$ is the magnitude of the displacement.
• $\theta$ is the angle between the force vector and the displacement vector.

The unit of work in the International System of Units (SI) is the joule (J). One joule is the work done when a force of one newton moves an object through a distance of one meter in its direction. In the CGS system, the unit of work is the erg. 1 J = 10⁷ ergs.

Cases for $\theta$:

• If $\theta = 0^\circ$ (force and displacement are in the same direction), $\cos \theta = 1$, so $W = |\vec{F}| |\vec{d}|$. Work done is positive. Example: Pushing a box across a floor in the direction of motion.
• If $\theta = 90^\circ$ (force is perpendicular to displacement), $\cos \theta = 0$, so $W = 0$. No work is done. Example: Carrying a bag horizontally at a constant speed – the gravitational force is downwards, but displacement is horizontal.
• If $\theta = 180^\circ$ (force and displacement are in opposite directions), $\cos \theta = -1$, so $W = -|\vec{F}| |\vec{d}|$. Work done is negative. Example: Friction acting on a moving object.
• If $0^\circ < \theta < 90^\circ$, $\cos \theta$ is positive, so $W$ is positive.
• If $90^\circ < \theta < 180^\circ$, $\cos \theta$ is negative, so $W$ is negative.

Work done by gravity: When an object moves vertically, the work done by gravity depends on the change in height. If an object of mass $m$ is raised by a height $h$, the force of gravity is $mg$ downwards. The displacement is $h$ upwards. The angle between the force of gravity and displacement is $180^\circ$. So, the work done by gravity is $W_g = -mgh$. If the object falls by a height $h$, the displacement is $h$ downwards, in the same direction as gravity. The angle is $0^\circ$. So, the work done by gravity is $W_g = +mgh$.

Work done by friction: Friction is a force that opposes motion. Therefore, the angle between the force of friction and the displacement is always $180^\circ$. This means that the work done by friction is always negative, which is why friction causes a loss of energy from a system.

Work Done by Variable Force

When the force acting on an object is not constant but varies with its position, we cannot use the simple formula $W = Fd \cos \theta$. In such cases, we need to use calculus to find the work done.

Consider a force $\vec{F}(x)$ that varies with position $x$. If the object moves from position $x_1$ to $x_2$ along the x-axis, the work done is the integral of the force component in the direction of motion over the displacement.

$W = \int_{x_1}^{x_2} F_x(x) dx$

If the force and displacement are in three dimensions, $\vec{F}(x, y, z)$ and the displacement is along a path $C$, the work done is given by the line integral:

$W = \int_{C} \vec{F} \cdot d\vec{r}$

Where $d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$ and $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$.

Geometrically, the work done by a variable force is equal to the area under the force-displacement graph. If we plot $F_x$ versus $x$, the area between the curve and the x-axis from $x_1$ to $x_2$ represents the work done.

Example: Work done by a spring force. The force exerted by an ideal spring is given by Hooke's Law: $F_s = -kx$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. This is a variable force. To find the work done in stretching or compressing a spring from $x_1$ to $x_2$:

$W_{spring} = \int_{x_1}^{x_2} (-kx) dx = -k \int_{x_1}^{x_2} x dx = -k \left[ \frac{x^2}{2} \right]_{x_1}^{x_2} = -\frac{1}{2}k(x_2^2 - x_1^2)$

If the spring is initially at its equilibrium position ($x_1 = 0$) and stretched to $x_2 = x$, the work done by the spring force is $W_{spring} = -\frac{1}{2}kx^2$. The work done by an external agent to stretch the spring is the negative of this, i.e., $W_{external} = +\frac{1}{2}kx^2$.

Kinetic Energy

Kinetic energy (KE) is the energy an object possesses due to its motion. An object in motion has the capacity to do work. The faster an object moves, the more kinetic energy it has.

The formula for kinetic energy is derived from the work-energy theorem. Consider an object of mass $m$ initially at rest, acted upon by a constant force $F$. The force causes the object to accelerate and move a distance $d$. According to Newton's second law, $F = ma$.

From kinematics, $v^2 = u^2 + 2ad$. If the initial velocity $u = 0$, then $v^2 = 2ad$, so $d = \frac{v^2}{2a}$.

The work done by the force is $W = Fd$. Substituting $F=ma$ and $d=\frac{v^2}{2a}$:

$W = (ma) \left(\frac{v^2}{2a}\right) = \frac{1}{2}mv^2$

This work done is converted into kinetic energy. Therefore, the kinetic energy of an object of mass $m$ moving with velocity $v$ is:

$KE = \frac{1}{2}mv^2$

The SI unit of kinetic energy is the joule (J), same as work.

Work-Energy Theorem: This theorem states that the net work done on an object is equal to the change in its kinetic energy.

$W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

This theorem is very useful because it relates work directly to the change in kinetic energy, without needing to know the details of the forces involved, only the net work done.

Example: A 1000 kg car is moving at 20 m/s. What is its kinetic energy?

$KE = \frac{1}{2}mv^2 = \frac{1}{2} (1000 \text{ kg}) (20 \text{ m/s})^2 = \frac{1}{2} (1000) (400) = 200,000 \text{ J} = 200 \text{ kJ}$.

If the driver applies the brakes and the car comes to a stop from 20 m/s, the work done by the braking force is equal to the change in kinetic energy.

$W_{braking} = \Delta KE = 0 - 200,000 \text{ J} = -200,000 \text{ J}$. The negative sign indicates that the braking force opposes the motion.

Potential Energy

Potential energy (PE) is the energy stored in an object or system by virtue of its position or configuration. It represents the potential to do work. There are different types of potential energy, but we will focus on gravitational and elastic potential energy.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. For an object near the Earth's surface, where the gravitational field is approximately uniform, the gravitational potential energy depends on its height above a reference level.

Let's set the reference level (where PE = 0) at the ground. If an object of mass $m$ is at a height $h$ above the ground, the force of gravity acting on it is $F_g = mg$ downwards. To lift the object to height $h$, an external agent must apply an upward force equal to $mg$ (assuming slow, steady lifting to avoid significant acceleration).

The work done by the external agent against gravity is $W_{external} = F_{applied} \times d = (mg) \times h = mgh$.

This work done is stored as gravitational potential energy in the object. So, the gravitational potential energy $U_g$ at height $h$ is:

$U_g = mgh$

Where:

• $m$ is the mass of the object.
• $g$ is the acceleration due to gravity (approximately 9.8 m/s² near Earth's surface).
• $h$ is the height above the chosen reference level.

The SI unit of potential energy is the joule (J).

Important Notes:

• Potential energy is always defined relative to a reference point. Changing the reference point changes the value of PE, but the difference in PE between two points remains the same.
• Gravitational potential energy is a property of the system (e.g., Earth-object system), not just the object itself.
• When an object falls from height $h_1$ to $h_2$, its potential energy decreases by $mg(h_1 - h_2)$. This decrease in PE is converted into kinetic energy, assuming no other forces do work.

Example: A 5 kg book is placed on a shelf 2 meters above the floor. What is its gravitational potential energy relative to the floor?

$U_g = mgh = (5 \text{ kg})(9.8 \text{ m/s}^2)(2 \text{ m}) = 98 \text{ J}$.

If the book falls to the floor ($h=0$), its potential energy becomes 0 J. The change in potential energy is $\Delta U_g = 0 - 98 \text{ J} = -98 \text{ J}$. This means 98 J of potential energy has been converted into kinetic energy (and possibly sound/heat due to impact).

Elastic Potential Energy

Elastic potential energy is the energy stored in a deformable object, such as a spring or a rubber band, when it is stretched or compressed from its equilibrium position.

For an ideal spring obeying Hooke's Law ($F_s = -kx$), the force exerted by the spring is proportional to its displacement from equilibrium. The work done by an external agent to stretch or compress the spring from an initial displacement $x_1$ to a final displacement $x_2$ is:

$W_{external} = \frac{1}{2}k(x_2^2 - x_1^2)$

This work done is stored as elastic potential energy ($U_e$) in the spring. If we take the equilibrium position ($x=0$) as the reference point where $U_e = 0$, then the elastic potential energy at a displacement $x$ is:

$U_e = \frac{1}{2}kx^2$

Where:

• $k$ is the spring constant (a measure of the stiffness of the spring).
• $x$ is the displacement from the equilibrium position.

The SI unit of elastic potential energy is the joule (J).

Example: A spring has a spring constant of 200 N/m.

• What is the elastic potential energy stored when it is stretched by 0.1 m from its equilibrium position?

$U_e = \frac{1}{2}kx^2 = \frac{1}{2}(200 \text{ N/m})(0.1 \text{ m})^2 = \frac{1}{2}(200)(0.01) = 1 \text{ J}$.

• What is the work done by the spring force when it is released from this stretched position back to equilibrium?

The work done by the spring force is $W_{spring} = -\Delta U_e = -(0 - 1 \text{ J}) = +1 \text{ J}$. The positive work done by the spring causes it to move back to equilibrium.

Conservation of Mechanical Energy

In the absence of non-conservative forces (like friction or air resistance), the total mechanical energy of a system remains constant. Mechanical energy is the sum of kinetic energy and potential energy.

$E = KE + PE$

If only conservative forces (like gravity or spring force) do work, then:

$E_{initial} = E_{final}$

$KE_i + PE_i = KE_f + PE_f$

Example: A freely falling object. Consider an object of mass $m$ dropped from a height $h$. Let's choose the ground as the reference for PE ($PE=0$).

• At height $h$ (initial):

$KE_i = 0$ (since it's dropped from rest)

$PE_i = mgh$

$E_i = 0 + mgh = mgh$

• Just before hitting the ground (final, height = 0):

Let the velocity be $v$.

$KE_f = \frac{1}{2}mv^2$

$PE_f = mg(0) = 0$

$E_f = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2$

• Conservation of Energy:

$E_i = E_f \implies mgh = \frac{1}{2}mv^2$

This gives $v^2 = 2gh$, or $v = \sqrt{2gh}$, which is consistent with kinematic equations. This shows that the potential energy lost is converted entirely into kinetic energy.

Example: A simple pendulum. When a pendulum swings, its energy continuously converts between kinetic and potential energy. At the highest points of the swing, the bob momentarily stops, so KE = 0 and PE is maximum. At the lowest point (equilibrium), PE is minimum (often taken as zero), and KE is maximum. The total mechanical energy remains constant if air resistance is ignored.

Power

Power is the rate at which work is done or energy is transferred. It tells us how quickly work is performed.

If an amount of work $W$ is done in time $t$, the average power $P_{avg}$ is given by:

$P_{avg} = \frac{W}{t}$

The SI unit of power is the watt (W). One watt is equal to one joule per second (1 W = 1 J/s). Another common unit is horsepower (hp), where 1 hp = 746 W.

Instantaneous Power: If the rate of work done is not constant, we can define instantaneous power as the derivative of work with respect to time:

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

Since $W = \vec{F} \cdot \vec{d}$, we can also express instantaneous power in terms of force and velocity:

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

If the force is constant, $P = \vec{F} \cdot \frac{d\vec{d}}{dt} = \vec{F} \cdot \vec{v}$.

$P = Fv \cos \theta$

Where $\theta$ is the angle between the force $\vec{F}$ and the velocity $\vec{v}$.

Example: A motor lifts a 10 kg object vertically by 5 meters in 2 seconds. What is the average power of the motor?

First, calculate the work done against gravity:

$W = mgh = (10 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m}) = 490 \text{ J}$.

Now, calculate the average power:

$P_{avg} = \frac{W}{t} = \frac{490 \text{ J}}{2 \text{ s}} = 245 \text{ W}$.

Example: A car engine can deliver a maximum power of 50 kW. If the car has a mass of 1000 kg and is moving at 20 m/s against a constant resistive force of 500 N, what is the magnitude of the accelerating force?

The power delivered by the engine is used to overcome the resistive force and to provide acceleration.

$P = F_{net} v$

$P = (F_{acceleration} - F_{resistance}) v$

$50,000 \text{ W} = (F_{acceleration} - 500 \text{ N}) (20 \text{ m/s})$

$\frac{50,000}{20} \text{ N} = F_{acceleration} - 500 \text{ N}$

$2500 \text{ N} = F_{acceleration} - 500 \text{ N}$

$F_{acceleration} = 2500 \text{ N} + 500 \text{ N} = 3000 \text{ N}$.