Chapter 4 numericals class 11 physics | physics ka safar

Physics ka Safar・2 minutes read

The text discusses various physics problems related to work, energy, and forces, illustrating calculations for different scenarios such as a man mowing grass, a falling raindrop, stacking bricks, and the motion of cars. It highlights the application of formulas to find work done, potential energy, and power, with specific numerical examples demonstrating the underlying principles of mechanics.

Insights

  • The work done by a man pushing a lawn mower can be calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), which incorporates the force applied, the distance covered, and the angle of application. In this case, with a force of 40 Newtons at a 20-degree angle over 20 meters, the work done amounts to approximately 750 Joules.
  • For a raindrop falling under the influence of gravity, the work done by gravity can be calculated as a positive value when moving downward, while the work done against friction is negative due to the opposing direction of the frictional force. Specifically, the gravitational work is about 0.0328 Joules, and the work done against friction is approximately -0.0328 Joules, illustrating the balance of forces acting on the raindrop.
  • When stacking 10 bricks, the work done increases with each brick placed higher, leading to a total work of about 0.44 Joules for all bricks combined. This demonstrates how the height and weight of objects influence the total work required in stacking operations, emphasizing the cumulative nature of work in physics.
  • The power output required by a car's engine to overcome friction and air resistance while traveling at 80 km/h is calculated to be approximately 17.78 kW. This highlights the significant energy demands placed on vehicles, particularly at higher speeds, and underscores the importance of understanding power and force relationships in real-world applications.

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Recent questions

  • What is the definition of work in physics?

    Work in physics is defined as the process of energy transfer that occurs when a force is applied to an object, causing it to move. Mathematically, work (W) is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved by the object in the direction of the force, and \( \theta \) is the angle between the force and the direction of motion. If the force is applied in the same direction as the movement, the work done is maximized. Work is measured in joules (J) in the International System of Units (SI), where one joule is equivalent to one newton-meter.

  • How do you calculate potential energy?

    Potential energy (PE) is calculated using the formula \( PE = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and \( h \) is the height above a reference point. This formula indicates that the potential energy of an object increases with its height and mass. For example, if an object with a mass of 10 kg is raised to a height of 5 meters, its potential energy would be calculated as \( PE = 10 \cdot 9.8 \cdot 5 \), resulting in 490 joules. This energy represents the work done against gravity to elevate the object to that height.

  • What is kinetic energy and how is it calculated?

    Kinetic energy (KE) is the energy that an object possesses due to its motion. It is calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. This formula shows that kinetic energy increases with the square of the velocity, meaning that even small increases in speed can result in significant increases in kinetic energy. For instance, if a car with a mass of 1000 kg is traveling at a speed of 20 m/s, its kinetic energy would be \( KE = \frac{1}{2} \cdot 1000 \cdot (20)^2 \), resulting in 200,000 joules. This energy is crucial in understanding the dynamics of moving objects and their interactions.

  • What is the work-energy principle?

    The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. This principle can be expressed mathematically as \( W = \Delta KE \), where \( W \) is the work done and \( \Delta KE \) is the change in kinetic energy. This means that if work is done on an object, it will either increase or decrease its kinetic energy depending on the direction of the force applied. For example, if a car is brought to a stop, the work done by the brakes is equal to the negative of the car's initial kinetic energy, resulting in a decrease in speed. This principle is fundamental in analyzing motion and energy transfer in various physical systems.

  • How is power defined in physics?

    Power in physics is defined as the rate at which work is done or energy is transferred over time. It is calculated using the formula \( P = \frac{W}{t} \), where \( P \) is power, \( W \) is the work done, and \( t \) is the time taken to do that work. Power is measured in watts (W) in the International System of Units (SI), where one watt is equivalent to one joule per second. For instance, if a machine does 1000 joules of work in 5 seconds, its power output would be \( P = \frac{1000}{5} = 200 \, \text{W} \). Understanding power is essential in various applications, from electrical devices to mechanical systems, as it indicates how quickly energy is being used or produced.

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Summary

00:00

Calculating Work in Various Physical Scenarios

  • A man pushes a lawn mower with a force of 40 Newtons at an angle of 20 degrees downward from the horizontal while cutting grass over a distance of 20 meters. The work done can be calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F = 40 \, \text{N} \), \( d = 20 \, \text{m} \), and \( \theta = 20^\circ \).
  • To find the work done, substitute the values into the formula: \( W = 40 \cdot 20 \cdot \cos(20^\circ) \). The result is approximately \( 750 \, \text{J} \) (or \( 7.5 \times 10^2 \, \text{J} \)).
  • The second numerical problem involves a raindrop falling vertically at a constant speed under the influence of gravity and friction, covering a distance of 100 meters. The mass of the raindrop is \( 3.35 \times 10^{-5} \, \text{kg} \).
  • The gravitational force acting on the raindrop can be calculated using \( F = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \). The work done by gravity is calculated using \( W = m \cdot g \cdot h \cdot \cos(0^\circ) \), resulting in \( W = 3.35 \times 10^{-5} \cdot 9.8 \cdot 100 \), which equals approximately \( 0.0328 \, \text{J} \).
  • The work done by friction is calculated similarly, but since the frictional force acts upward while the displacement is downward, the angle is \( 180^\circ \). Thus, \( W = m \cdot g \cdot h \cdot \cos(180^\circ) \), resulting in \( W = -3.35 \times 10^{-5} \cdot 9.8 \cdot 100 \), which equals approximately \( -0.0328 \, \text{J} \).
  • In the third numerical problem, the task is to determine the work required to stack 10 bricks, each with a thickness of 6 cm (0.06 m) and a mass of 1.5 kg, on top of each other. The work done to place the first brick is zero, as it is already on the table.
  • For each subsequent brick, the work done increases based on the height at which it is placed. The work for the second brick is \( W = m \cdot g \cdot h \), for the third brick \( W = 2 \cdot m \cdot g \cdot h \), and so on, up to the tenth brick, which requires \( W = 9 \cdot m \cdot g \cdot h \).
  • The total work done to stack all 10 bricks is the sum of the individual works: \( W_{\text{total}} = 0 + 1 \cdot m \cdot g \cdot h + 2 \cdot m \cdot g \cdot h + ... + 9 \cdot m \cdot g \cdot h \), which simplifies to \( W_{\text{total}} = 45 \cdot m \cdot g \cdot h \).
  • Substituting the values \( m = 1.5 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 0.06 \, \text{m} \) gives \( W_{\text{total}} = 45 \cdot 1.5 \cdot 9.8 \cdot 0.06 \), resulting in approximately \( 0.44 \, \text{J} \).
  • The fourth numerical problem involves a car with a mass of 800 kg traveling at 54 km/h (converted to 15 m/s) that comes to rest over a distance of 60 meters. The average retarding force can be calculated using the work-energy principle, where the work done equals the change in kinetic energy. The initial kinetic energy is \( KE = \frac{1}{2} m v^2 \), and the work done to stop the car is equal to the negative of this kinetic energy.

16:16

Energy and Force Calculations in Motion

  • The problem begins with calculating the returning force, which is determined to be approximately -100 Newton, indicating it acts in the opposite direction to the motion of the car, effectively stopping it.
  • The kinetic energy of the car decreases as it slows down from an initial velocity of 54 km/h due to the application of brakes, confirming that kinetic energy is inversely related to speed.
  • A numerical example involves a 1000 kg automobile released from a height of 10 meters on a 100-meter incline, with a frictional returning force of 480 Newton, requiring the calculation of its speed at the bottom of the incline.
  • The gravitational potential energy at the top is calculated using the formula PE = mgh, where m = 1000 kg, g = 9.8 m/s², and h = 10 m, resulting in an initial potential energy of 98,000 Joules.
  • As the car descends, some potential energy is converted into kinetic energy while some is lost to friction, leading to the equation: KE + Work done against friction = Initial PE.
  • The final velocity (v) at the bottom is derived from the equation: 1/2 mv² = Initial PE - Work done against friction, leading to a calculated speed of 10 m/s at the bottom of the incline.
  • Another numerical problem involves pumping 100 cubic meters of water to a height of 10 meters in 20 minutes, requiring the calculation of the increase in potential energy and the power delivered by the pump.
  • The mass of the water is calculated as 100,000 kg (using density = 1000 kg/m³), and the increase in potential energy is found to be 980,000 Joules.
  • The power delivered by the pump is calculated by dividing the potential energy by the time taken (1200 seconds), resulting in a power output of approximately 816.67 Watts or 0.82 kW.
  • A final example discusses a car requiring 400 Newtons of force to overcome friction and air resistance while traveling at 80 km/h, leading to the calculation of the power needed by the engine, which is derived from the formula P = Fv, resulting in a power output of approximately 17.78 kW.

32:07

Physics of Speed Energy and Force Calculations

  • To convert speed from kilometers per hour to meters per second, multiply the speed by 1000 (to convert kilometers to meters) and then divide by 3600 (the number of seconds in an hour). For example, 80 km/h equals approximately 22.22 m/s.
  • The power required for an engine operating at this speed is calculated to be 8.9 kilowatts, indicating the engine's power output needed for this velocity.
  • To determine the force required to accelerate an electron from rest to a speed of 2 x 10^7 m/s over a distance of 0.05 meters (5 centimeters), use the work-energy principle, where the work done equals the change in kinetic energy.
  • The formula for work done (W) is W = F * d, where F is force and d is distance. The change in kinetic energy is given by ΔKE = 0.5 * m * (v^2 - u^2), where m is mass, v is final velocity, and u is initial velocity.
  • The calculated force required to accelerate the electron is approximately 3.6 x 10^-15 Newtons.
  • A driver weighing 750 Newtons jumps from a height of 10 meters to a height of 5 meters above a pool. The speed at 5 meters can be found using the conservation of mechanical energy, where the loss in potential energy equals the gain in kinetic energy.
  • The loss in potential energy when descending from 10 meters to 5 meters is calculated as mgh, resulting in a speed of approximately 9.9 m/s when reaching the halfway point.
  • A child sliding down a frictionless slide of height 4 meters starts from rest. The speed at the bottom can be calculated using the formula v = √(2gh), resulting in a speed of approximately 8.8 m/s.
  • If the child reaches the bottom with a speed of 6 m/s due to friction, the percentage loss in energy due to friction can be calculated using the formula: (loss in energy / total energy) * 100.
  • The total energy at the top is 39.2 Joules, and the loss in energy due to friction is 21.2 Joules, leading to a percentage loss of approximately 54% due to friction.
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