Work, Energy, and Simple MachinesClass 9 Science Notes

Work, Energy, and Simple Machines

In physics, the concepts of work, energy, and power provide a powerful alternative to Newton's laws for analyzing the motion of objects, especially when forces are complex or change over time. At the heart of these ideas is energy, which is defined as the capacity to do work.

Work Done by a Constant Force

In science, work is done on an object only when a force causes it to move through a distance. The amount of work done depends on both the size of the force and the distance the object moves.

For a constant force, the work done is defined as the product of the force and the displacement in the direction of that force.

Work done on an object by a constant force = force applied × displacement in the direction of the force

If a constant force $F$ acts on an object, causing it to have a displacement $s$ in the same direction as the force, the work done $W$ is given by the formula: $W = F \times s$

The SI unit of work is the joule, represented by the symbol J. One joule is the amount of work done when a force of 1 newton displaces an object by 1 metre in the direction of the force. $1 \text{ J} = 1 \text{ N} \times 1 \text{ m}$ Since $1 \text{ N} = 1 \text{ kg m s}^{-2}$, the joule can also be expressed in base SI units: $1 \text{ J} = 1 \text{ kg m s}^{-2} \times 1 \text{ m} = 1 \text{ kg m}^2 \text{ s}^{-2}$

On a graph plotting force versus displacement, the work done is equal to the area under the curve.

When is work done equal to zero?

Even if a force is applied, no scientific work is done under the following conditions:

• No displacement: If the object does not move ($s=0$), the work done is zero. For example, pushing against a rigid wall requires effort and makes you feel tired, but since the wall doesn't move, no work is done on the wall. The feeling of tiredness comes from your muscles using internal energy.
• No force: If there is no force acting on the object ($F=0$), no work can be done.
• Force perpendicular to displacement: If the force acts in a direction perpendicular (at a 90° angle) to the object's displacement, the work done by that specific force is zero. This is because there is no displacement in the direction of the force.

Positive and negative work done

The work done by a force can be positive, negative, or zero, depending on the angle between the force and the displacement.

• Positive work is done when the force and displacement are in the same direction. The force helps to increase the object's energy. For example, when you push a wheelchair, the force you apply and the wheelchair's movement are in the same direction.
• Negative work is done when the force and displacement are in opposite directions. The force acts to decrease the object's energy. For example, when a goalkeeper stops a football, the force applied by her hands is opposite to the ball's motion.

Solution

When the girl lifts the dumbbell up, the force she applies is upwards, and the displacement is also upwards. Since the force and displacement are in the same direction, she does positive work on the dumbbell.

When she lowers the dumbbell, she still applies an upward force to control its descent, but the displacement is downwards. Since the force and displacement are in opposite directions, she does negative work on the dumbbell.

Given

• Force applied by goalkeeper, $F = 200 \text{ N}$
• Displacement of the ball, $s = 15 \text{ cm} = 0.15 \text{ m}$

To Find

• Work done by the goalkeeper on the ball.

Formula

$W = F \times s$

Solution

The goalkeeper applies a force opposite to the direction of the ball's motion. Therefore, the work done on the ball is negative. We can represent the displacement in the direction of the force as negative.

$W = 200 \text{ N} \times (-0.15 \text{ m})$ $W = -30 \text{ J}$

Final Answer The goalkeeper did $-30 \text{ J}$ of work on the ball.

The Work-Energy Theorem

When work is done on an object, its energy changes. This fundamental connection is described by the work-energy theorem, which states:

Work done on an object = Change in its energy

This powerful theorem means that if you do positive work on an object, it gains energy. If you do negative work on it, it loses energy. This relationship holds true even for changing forces and complex systems.

Solution

1. Striker hits White Coin: The moving striker applies a force on the stationary white coin, causing it to move. The striker does positive work on the white coin, transferring energy to it and increasing its energy. According to Newton's third law, the white coin exerts an equal and opposite force on the striker, doing negative work on the striker and decreasing its energy.
2. White Coin hits Black Coin: The now-moving white coin applies a force on the black coin. The white coin does positive work on the black coin, increasing its energy and setting it in motion. Simultaneously, the black coin does negative work on the white coin, decreasing its energy.

Forms of Energy

Energy exists in many forms, and it can be converted from one form to another. Some common forms include:

• Mechanical Energy: Energy of motion and position.
• Electrical Energy: Energy from the flow of electric charge.
• Light Energy: Energy carried by electromagnetic waves.
• Thermal Energy: Heat energy related to the temperature of an object.
• Chemical Energy: Energy stored in the bonds of chemical compounds (e.g., food, batteries).
• Sound Energy: Energy carried by vibrations through a medium.

For example, a light bulb converts electrical energy into light and thermal energy. Our bodies convert chemical energy from food into mechanical energy for movement.

Mechanical Energy

Mechanical energy is the energy an object possesses due to its motion or its position. It is the sum of two types of energy: kinetic energy and potential energy.

Kinetic energy

The energy an object has because of its motion is called kinetic energy. Any moving object, like a rolling ball or a moving car, has kinetic energy.

To find an expression for kinetic energy, we can use the work-energy theorem. The work done by a net force in accelerating an object from an initial velocity $u$ to a final velocity $v$ is: $W = \frac{1}{2} m(v^2 - u^2)$ This work done equals the change in the object's kinetic energy. If the object starts from rest ($u=0$), the work done gives the object its final kinetic energy.

The formula for the kinetic energy ($K$) of an object of mass $m$ moving with a velocity $v$ is: $K = \frac{1}{2} m v^2$

• Kinetic energy is a scalar quantity (it has magnitude but no direction).
• Its SI unit is the joule (J).
• If positive work is done on an object, its velocity and kinetic energy increase.
• If negative work is done, its velocity and kinetic energy decrease.

Solution

Let the initial kinetic energy be $K_{initial} = \frac{1}{2}mv^2$.

If the velocity doubles, the new velocity is $2v$. The new kinetic energy will be: $K_{new} = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \left(\frac{1}{2}mv^2\right) = 4 \times K_{initial}$

Final Answer If the velocity doubles, the kinetic energy becomes 4 times its original value.

Given

• Mass of the ball, $m = 0.2 \text{ kg}$
• Velocity of the ball, $v = 154.8 \text{ km h}^{-1}$

To Find

• Kinetic energy of the ball, $K$.

Formula

$K = \frac{1}{2} m v^2$

Solution

First, convert the velocity to SI units ($\text{m s}^{-1}$). $v = 154.8 \frac{\text{km}}{\text{h}} = 154.8 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 43 \text{ m s}^{-1}$ Now, substitute the values into the kinetic energy formula: $K = \frac{1}{2} \times 0.2 \text{ kg} \times (43 \text{ m s}^{-1})^2$ $K = 0.1 \times 1849 \text{ J} = 184.9 \text{ J}$

Final Answer The kinetic energy of the ball is $184.9 \text{ J}$.

Given

• Mass of the aircraft, $m = 15000 \text{ kg}$
• Final velocity, $v_{final} = 0 \text{ m s}^{-1}$
• Force exerted by the wire, $F = 367500 \text{ N}$ (backward)
• Displacement, $s = 100 \text{ m}$

To Find

• Initial velocity of the aircraft, $v$.

Formula

Work-energy theorem: Work done = Change in kinetic energy $W = K_{final} - K_{initial}$ $F \times s = \frac{1}{2}mv_{final}^2 - \frac{1}{2}mv^2$

Solution

The force exerted by the wire is backward, opposite to the displacement. Therefore, the work done by the wire is negative. $W = -F \times s = -367500 \text{ N} \times 100 \text{ m} = -36,750,000 \text{ J}$ The change in kinetic energy is: $\Delta K = K_{final} - K_{initial} = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(15000 \text{ kg})v^2$ According to the work-energy theorem: $-36,750,000 \text{ J} = -\frac{1}{2}(15000 \text{ kg})v^2$ $36,750,000 = 7500 \times v^2$ $v^2 = \frac{36,750,000}{7500} = 4900 \text{ m}^2\text{s}^{-2}$ $v = \sqrt{4900} = 70 \text{ m s}^{-1}$ To convert this to $\text{km h}^{-1}$: $v = 70 \frac{\text{m}}{\text{s}} \times \frac{3600 \text{ s}}{1 \text{ h}} \times \frac{1 \text{ km}}{1000 \text{ m}} = 252 \text{ km h}^{-1}$

Final Answer The velocity of the aircraft was $70 \text{ m s}^{-1}$ (or $252 \text{ km h}^{-1}$).

Potential energy

The energy that is stored in an object or a system due to its position, shape, or configuration is called potential energy. This stored energy has the "potential" to be converted into other forms of energy, such as kinetic energy.

Examples:

• A stretched rubber band has potential energy due to its deformation. When released, this energy is converted into the kinetic energy of a projectile.
• A compressed spring has potential energy.
• A system of objects interacting through forces like gravity, magnetism, or electricity can store potential energy due to the relative positions of the objects.

Gravitational Potential Energy

An object has gravitational potential energy due to its position in a gravitational field. When you lift an object of mass $m$ to a height $h$ near the Earth's surface, you do work against the force of gravity ($mg$). This work is stored in the object as gravitational potential energy.

The work done in lifting the object is: $W = \text{force} \times \text{displacement} = mg \times h$ According to the work-energy theorem, this work done is equal to the gain in potential energy ($U$). $U = mgh$

• The SI unit for potential energy is the joule (J).
• Gravitational potential energy is relative. We usually define the ground or the lowest point in a problem as the position of zero potential energy ($h=0$).
• The higher an object is, the greater its gravitational potential energy.

Given

• Mass of the ball, $m = 200 \text{ g} = 0.2 \text{ kg}$
• Height, $h = 10 \text{ m}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

To Find

• Potential energy of the ball, $U$.

Formula

$U = mgh$

Solution

Substitute the given values into the formula: $U = 0.2 \text{ kg} \times 10 \text{ m s}^{-2} \times 10 \text{ m}$ $U = 20 \text{ J}$

Final Answer The ball has $20 \text{ J}$ of potential energy at its maximum height.

Conservation of mechanical energy

The conservation of mechanical energy is a fundamental principle which states that if an object is only under the influence of gravitational force (with no other external forces like friction or air resistance), its total mechanical energy remains constant.

Total Mechanical Energy = Kinetic Energy + Potential Energy $E = K + U = \text{constant}$

As a freely falling object moves downwards, its height decreases, so its potential energy is converted into kinetic energy, and its speed increases. The loss in potential energy is exactly equal to the gain in kinetic energy.

In a real pendulum, air resistance and friction at the pivot cause some mechanical energy to be lost (converted to heat), so the pendulum eventually stops.

Given

• Initial height = $h$
• Initial velocity (at the top), $u = 0$
• Final height = $0$

To Find

• Final velocity at the bottom, $v$.

Formula

Conservation of Mechanical Energy: $E_{top} = E_{bottom}$ $(K + U)_{top} = (K + U)_{bottom}$ $\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv^2 + mg(0)$

Solution

At the top of the slide, the child is at rest ($u=0$), so kinetic energy is zero. Potential energy is $mgh$. $E_{top} = 0 + mgh = mgh$ At the bottom of the slide, the height is zero, so potential energy is zero. The kinetic energy is $\frac{1}{2}mv^2$. $E_{bottom} = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2$ By the conservation of energy: $mgh = \frac{1}{2}mv^2$ The mass $m$ cancels out from both sides: $gh = \frac{1}{2}v^2$ $v^2 = 2gh$ $v = \sqrt{2gh}$

Final Answer The velocity of the child at the bottom is $v = \sqrt{2gh}$. This result shows that the final velocity depends only on the height of the slide, not on the child's mass or the shape of the slide (assuming no friction).

Given

• Mass of the truck, $m = 10000 \text{ kg}$
• Initial velocity, $v_i = 72 \text{ km h}^{-1} = 20 \text{ m s}^{-1}$
• Final velocity, $v_f = 0 \text{ m s}^{-1}$
• Resistive force from sand, $F_{sand} = 50000 \text{ N}$
• Gravitational acceleration, $g = 10 \text{ m s}^{-2}$

To Find

• Minimum length of the ramp, $d$.

Formula

Work-Energy Theorem: Work done by all forces = Change in Kinetic Energy $W_{net} = \Delta K$ The net work is the sum of work done by gravity and work done by the sand.

Solution

Let the truck travel a distance $d$ along the ramp. From the hint, the vertical height gained is $h = d/2$.

The initial energy of the truck is purely kinetic (we set initial potential energy to zero). $K_{initial} = \frac{1}{2}mv_i^2 = \frac{1}{2} \times 10000 \text{ kg} \times (20 \text{ m s}^{-1})^2 = 2,000,000 \text{ J}$ The final energy of the truck is purely potential, as it comes to rest ($v_f=0$). $U_{final} = mgh = 10000 \text{ kg} \times 10 \text{ m s}^{-2} \times \frac{d}{2} = 50000d \text{ J}$ The sand does negative work on the truck because its force opposes the motion. $W_{sand} = -F_{sand} \times d = -50000d \text{ J}$ The work-energy theorem states that the work done by non-conservative forces (like sand friction) equals the change in the total mechanical energy. $W_{sand} = E_{final} - E_{initial}$ $W_{sand} = (K_{final} + U_{final}) - (K_{initial} + U_{initial})$ $-50000d = (0 + 50000d) - (2,000,000 + 0)$ $-50000d = 50000d - 2,000,000$ $2,000,000 = 50000d + 50000d$ $2,000,000 = 100000d$ $d = \frac{2,000,000}{100,000} = 20 \text{ m}$

Final Answer The minimum length of the ramp required is $20 \text{ m}$.

Power

While work tells us about the energy transferred, it doesn't say how quickly it happened. Power is the physical quantity that describes how fast work is done or how fast energy is transferred.

Power is defined as the rate at which work is done. $P = \frac{W}{t}$ Where $P$ is power, $W$ is the work done, and $t$ is the time taken.

The SI unit of power is the watt (W), named after James Watt. One watt is equal to one joule of work done per second. $1 \text{ W} = 1 \text{ J s}^{-1}$

Doing the same amount of work in less time requires more power.

Given

• Mass, $m = 75 \text{ kg}$
• Height, $h = 2 \text{ m}$
• Time, $t = 5 \text{ s}$
• $g = 10 \text{ m s}^{-2}$

To Find

• Power required, $P$.

Formula

$P = \frac{W}{t} \quad \text{and} \quad W = mgh$

Solution

First, calculate the work done to lift the mass. This work is equal to the gain in potential energy. $W = mgh = 75 \text{ kg} \times 10 \text{ m s}^{-2} \times 2 \text{ m} = 1500 \text{ J}$ Now, calculate the power. $P = \frac{W}{t} = \frac{1500 \text{ J}}{5 \text{ s}} = 300 \text{ W}$

Final Answer The power required is $300 \text{ W}$.

Given

• Mass, $m = 1000 \text{ kg}$
• Initial velocity, $u = 0 \text{ m s}^{-1}$
• Final velocity, $v = 72 \text{ km h}^{-1} = 20 \text{ m s}^{-1}$
• Time, $t = 10 \text{ s}$

To Find

• Power of the engine, $P$.

Formula

$P = \frac{W}{t} \quad \text{and} \quad W = \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Solution

First, calculate the work done by the engine. This work is equal to the change in the car's kinetic energy. $W = \frac{1}{2} \times 1000 \text{ kg} \times (20 \text{ m s}^{-1})^2 - 0$ $W = 500 \times 400 = 200,000 \text{ J}$ Now, calculate the average power. $P = \frac{W}{t} = \frac{200,000 \text{ J}}{10 \text{ s}} = 20,000 \text{ W}$

Final Answer The power of the engine is $20,000 \text{ W}$ (or $20 \text{ kW}$).

Simple Machines

Simple machines are devices that help us do work by changing the magnitude or direction of the force we apply. They make tasks feel easier, but they do not reduce the total amount of work done (ignoring friction).

• Effort: The force we apply to the machine.
• Load: The force that the machine must overcome (e.g., the weight of an object).

The effectiveness of a machine in multiplying force is described by its mechanical advantage. $\text{mechanical advantage} = \frac{\text{load}}{\text{effort}}$

Pulley

A pulley is a wheel with a groove that guides a rope.

• A fixed pulley is attached to a support and does not move. It changes the direction of the effort. For example, it allows you to pull down on a rope to lift a load up. This is often more convenient than pulling upwards.
• For an ideal fixed pulley, the effort is equal to the load, so its mechanical advantage is 1.

Inclined plane

An inclined plane, such as a ramp, is a flat supporting surface tilted at an angle. It is used to move a heavy load to a higher or lower level.

• An inclined plane reduces the effort needed to lift an object compared to lifting it vertically.
• However, the distance over which the effort must be applied is increased.
• The mechanical advantage of an ideal (frictionless) inclined plane is the ratio of the length of the plane ($L$) to the vertical height ($h$). $\text{mechanical advantage} = \frac{L}{h}$ Since $L$ is always greater than $h$, the mechanical advantage of an inclined plane is always greater than 1. A longer, shallower ramp requires less effort.

Given

• Height of the ramp, $h = 30 \text{ cm}$
• Width (base) of the ramp = $40 \text{ cm}$

To Find

• Mechanical advantage of the ramp.

Formula

$\text{mechanical advantage} = \frac{L}{h}$

Solution

The ramp forms a right-angled triangle with height 30 cm and base 40 cm. We need to find the length of the ramp ($L$), which is the hypotenuse. Using the Pythagorean theorem: $L^2 = h^2 + \text{width}^2$ $L^2 = (30 \text{ cm})^2 + (40 \text{ cm})^2 = 900 + 1600 = 2500 \text{ cm}^2$ $L = \sqrt{2500} = 50 \text{ cm}$ Now, calculate the mechanical advantage: $\text{mechanical advantage} = \frac{L}{h} = \frac{50 \text{ cm}}{30 \text{ cm}} \approx 1.67$

Final Answer The mechanical advantage of the ramp is approximately $1.67$.

Lever

A lever is a rigid bar that rotates about a fixed point called the fulcrum. Levers are used to multiply force, making it possible to lift heavy loads with a small effort. A lever has three main parts:

• Fulcrum: The pivot point.
• Load: The force to be overcome.
• Effort: The force applied.

The distance from the fulcrum to the load is the load arm, and the distance from the fulcrum to the effort is the effort arm.

For a lever in balance, the principle of moments states: effort × effort arm = load × load arm

The mechanical advantage of a lever is given by: $\text{mechanical advantage} = \frac{\text{load}}{\text{effort}} = \frac{\text{effort arm}}{\text{load arm}}$ By making the effort arm longer than the load arm, a small effort can be used to overcome a large load, resulting in a mechanical advantage greater than 1.

Given

• Mass of child 1, $m_1 = 15 \text{ kg}$
• Mass of child 2, $m_2 = 30 \text{ kg}$
• Distances from fulcrum C: $\text{AC} = 2 \text{ m}$, $\text{BC} = 1 \text{ m}$, $\text{DC} = 1 \text{ m}$, $\text{EC} = 2 \text{ m}$

To Find

• The seats where the children should sit to balance the seesaw.

Formula

For balance: effort × effort arm = load × load arm

Solution

Let the 15 kg child be the "effort" and the 30 kg child be the "load". Let's place the 15 kg child on seat A, which has an effort arm of 2 m. The "effort moment" is $15 \text{ kg} \times 2 \text{ m} = 30 \text{ kg} \cdot \text{m}$.

To balance this, the "load moment" from the 30 kg child must be equal. Let the distance for the 30 kg child be $L$. $30 \text{ kg} \times L = 30 \text{ kg} \cdot \text{m}$ $L = \frac{30}{30} = 1 \text{ m}$ The 30 kg child must sit at a distance of 1 m from the fulcrum. This corresponds to seat D.

Final Answer The 15 kg child should sit on seat A (or E), and the 30 kg child should sit on seat D (or B).

Classes of Levers

Levers are categorized into three classes based on the relative positions of the fulcrum, effort, and load.

Bridging Science and Society: The Gharat

In the Himalayan region, traditional water mills called gharat or panchakki are a perfect example of energy transformation.

1. Water at a high elevation has gravitational potential energy.
2. As the water flows down a pipe, this potential energy is converted into kinetic energy.
3. The kinetic energy of the flowing water turns a wheel, converting it into rotational mechanical energy.
4. This rotating wheel is connected to a grinding stone, which does the work of grinding grain.

This same principle of converting potential energy to kinetic energy to rotational energy is used on a much larger scale in modern hydroelectric dams to generate electricity.