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Work, Energy, and

🎓 Class 9📖 Exploration📖 8 notes🧠 15 Q&A⏱️ ~12 min

Work, Energy, andStudy Notes

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Work Done by a Constant Force

Explanation

Work Done by a Constant Force

Work is a fundamental concept in physics that describes the effect of a force causing displacement of an object. Scientifically, work is said to be done when a force is applied on an object and the object moves in the direction of the applied force. Consider the example of lifting a wheat bag of mass 5 kg from the floor to a height of 1 meter. The gravitational force acting downward on the bag is mg, where m is the mass and g is the acceleration due to gravity. To lift the bag slowly, an upward force equal to mg must be applied. The bag is displaced upward by 1 meter in the direction of the applied force, so work is done. If multiple bags are lifted one after another to the same height, the total work done is proportional to the number of bags. Similarly, lifting a single bag to a greater height increases the work done proportionally. This leads to the scientific definition of work done by a constant force: work done = force applied × displacement in the direction of the force. This definition applies regardless of the direction of force and displacement, whether vertical, horizontal, or any other direction. For instance, if a constant force F acts on an object causing displacement s in the direction of the force, the work done W is W = F × s. It is important to specify both the force doing the work and the object on which work is done. The SI unit of work is the joule (J), where 1 joule equals 1 newton of force causing displacement of 1 meter in the force direction. Mathematically, 1 J = 1 N × 1 m = 1 kg·m²/s². When the force is not constant, the work done can be calculated as the area under the force-displacement graph between initial and final positions. Work done is zero if either the force is zero or the displacement is zero. For example, pushing a rigid wall with no displacement results in zero work done, even though effort is exerted. Similarly, if the force is perpendicular to displacement, such as carrying a box horizontally while applying upward force, the work done by that force is zero. Work done can be positive or negative depending on the relative directions of force and displacement. Positive work occurs when force and displacement are in the same direction, such as pushing a wheelchair forward. Negative work occurs when force and displacement are opposite, such as a goalkeeper applying force opposite to the motion of a ball to stop it. This section lays the foundation for understanding how forces cause energy changes in objects through work.

  • Work is done when a force causes displacement in the direction of the force.
  • Work done = force × displacement in the direction of force.
  • SI unit of work is joule (J), where 1 J = 1 N × 1 m.
  • Work is zero if force is zero or displacement is zero.
  • Work done can be positive or negative depending on force and displacement directions.
  • Work done equals the area under force-displacement graph if force varies.
  • 📌 Work: Product of force and displacement in the direction of force.
  • 📌 Force: A push or pull acting on an object.
  • 📌 Displacement: The distance moved in a specific direction.

The Work-Energy Theorem

Explanation

The Work-Energy Theorem

The work-energy theorem establishes a fundamental relationship between the work done on an object and the change in its energy. When a force causes displacement, work is done on the object, and this work results in a change in the object's energy. Consider everyday examples such as a fielder throwing a cricket ball towards the wickets or a flowerpot raised to a height and then dropped. The moving ball or falling pot has the capacity to do work on other objects, such as knocking down the wickets or damaging an object below. This capacity to do work is called energy. The ball gains energy from the work done by the fielder in throwing it, and the pot gains energy from the work done in raising it to a height. Positive work done on an object increases its energy, which can later be transferred to other objects through forces causing motion. Mathematically, the work-energy theorem is stated as: work done on an object = change in its energy. This theorem applies even when forces are not constant or when multiple objects interact. It provides a powerful tool to analyze motion and energy changes without detailed force analysis. The SI unit of energy is the joule (J), the same as work. Energy can be transferred not only mechanically but also as heat, radiation, sound, or electrical energy. For example, the Sun's energy reaches Earth as radiation, and energy flows from hot to cold objects as heat. James Prescott Joule, after whom the unit joule is named, studied the relationship between mechanical and thermal energy, establishing the principle of energy conservation. An example in a carrom game demonstrates the work-energy theorem: the striker does positive work on the white coin, increasing its energy; the white coin does negative work on the striker, decreasing its energy; similarly, energy transfers continue through collisions. Understanding the work-energy theorem helps explain many physical phenomena and solve problems involving forces and motion more simply.

  • Work done on an object equals the change in its energy (work-energy theorem).
  • Energy is the capacity to do work.
  • Positive work increases an object's energy; negative work decreases it.
  • Energy transfer can occur mechanically, as heat, radiation, or sound.
  • The SI unit of energy is the joule (J), same as work.
  • James Prescott Joule established the relation between mechanical and thermal energy.
  • 📌 Energy: Capacity to do work.
  • 📌 Work-energy theorem: Work done on an object equals change in its energy.
  • 📌 Positive work: Work done when force and displacement are in the same direction.

Forms of Energy

Concept

Forms of Energy

Energy exists in various forms beyond mechanical energy, which is associated with motion and position. Examples of different forms of energy include electrical, chemical, thermal, sound, light, and nuclear energy. Energy can be transformed from one

Practice QuestionsWork, Energy, and

Includes NCERT exercise questions with answers

Q1.Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.

Answer:

The energy required to raise a flag depends on the height of the flagpole and the weight (mass) of the flag. The work done is equal to the gravitational potential energy gained by the flag, which is mgh (mass × gravity × height). Raising the flag slowly or quickly does not change the amount of work done because work depends only on force and displacement, not on time. If the speed of raising the flag is doubled, the power requirement doubles because power is work done per unit time. Doubling speed halves the time, so power = work/time doubles.

Explanation:

Work done = mgh, independent of speed. Power = Work / time. Doubling speed halves time, so power doubles.

EasyNCERT
Q2.A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.

Answer:

First day total mass = 60 + 100 = 160 kg Second day total mass = 60 + 40 + 100 = 200 kg Kinetic energy on day 1 = (1/2) × 160 × v² = 80 v² Kinetic energy on day 2 = (1/2) × 200 × v² = 100 v² Since energy transfer is from fuel, fuel used ∝ kinetic energy. Ratio of fuel used on day 2 to day 1 = 100 v² / 80 v² = 5/4 = 1.25 So, the fuel used on the second day is 1.25 times that on the first day.

Explanation:

Calculate kinetic energies for both days and find ratio. Fuel used ∝ kinetic energy as per assumption.

MediumNCERT
Q3.On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.

Answer:

Let the weight of the child be W and the adult be 2W. For balance, torque by child = torque by adult W × d1 = 2W × d2 => d1 = 2 d2 So, the child sits twice as far from the fulcrum as the adult. Draw a seesaw with fulcrum in the center, child on one side at distance d1, adult on other side at distance d2 = d1/2. This shows balance despite adult being heavier.

Explanation:

Balance condition: torque by child = torque by adult. Since adult is twice heavier, child sits twice farther.

EasyNCERT
Q4.A ball of mass 2 kg is thrown up with a velocity of 20 m/s. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m/s²).

Answer:

(i) Work done by gravity during upward motion is negative because gravity acts downward while displacement is upward. Work done by gravity during downward motion is positive because gravity and displacement are in the same direction. (ii) Initial kinetic energy = (1/2) m v² = 0.5 × 2 × 20² = 400 J Potential energy at max height = mgh = 2 × 10 × 19.4 = 388 J Work done by air resistance = Initial KE - Potential Energy = 400 - 388 = 12 J (energy lost due to air resistance) So, air resistance did -12 J of work (negative work).

Explanation:

Sign of work depends on direction of force and displacement. Energy lost due to air resistance = difference between initial KE and PE at max height.

MediumNCERT
Q5.A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?

Answer:

(i) At 0 m, KE = 180 J KE = (1/2) m v² => 180 = 0.5 × 10 × v² => v² = 36 => v = 6 m/s (ii) From the graph (Fig. 7.37), work done by force from 0 to 4 m is area under force-displacement curve. Calculate work done (area under curve): Assuming the graph shows force varying, calculate net work done (say W). Final KE = Initial KE + Work done From the figure (not provided here), suppose work done is 60 J (example). Final KE = 180 + 60 = 240 J Then, final speed v = sqrt(2 × KE / m) = sqrt(2 × 240 / 10) = sqrt(48) ≈ 6.93 m/s Negative acceleration occurs where force is opposite to motion (force negative). From graph, if force dips below zero, block decelerates in that region.

Explanation:

Calculate initial speed from KE. Calculate work done from force-displacement graph. Add work done to initial KE to get final KE. Calculate final speed. Check graph for negative force indicating negative acceleration.

HardNCERT
Q6.The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?

Answer:

Given: Height on Earth, h_E = 8 m Acceleration due to gravity on Earth, g_E = 9.8 m/s² (approx 10 m/s²) Acceleration due to gravity on Moon, g_M = g_E / 6 ≈ 10/6 ≈ 1.67 m/s² Using the formula for maximum height: h = v² / (2g) From Earth data: 8 = v² / (2 × 10) => v² = 8 × 20 = 160 On Moon, height h_M = v² / (2 g_M) = 160 / (2 × 1.67) = 160 / 3.34 ≈ 47.9 m So, the ball will travel approximately 48 m high on the Moon.

Explanation:

Calculate initial velocity from Earth data. Use same velocity and Moon gravity to find height.

MediumNCERT
Q7.A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?

Answer:

(i) Between A and B, the car moves at constant speed (no acceleration) as the velocity is constant. (ii) Kinetic energy at A: Mass m = 1000 kg Velocity v (from graph at A) = say 20 m/s (assumed from typical graph) KE = (1/2) m v² = 0.5 × 1000 × 20² = 0.5 × 1000 × 400 = 200,000 J (iii) Work done by brakes = change in kinetic energy = KE at B - KE at C At C, car stops, KE = 0 So, work done by brakes = -200,000 J (negative work, removing energy) (iv) The kinetic energy is transformed into heat energy due to friction between brake pads and wheels, and sound energy.

Explanation:

Car moves at constant speed A to B. Calculate KE at A. Work done by brakes equals loss in KE. Energy converts to heat and sound.

MediumNCERT
Q8.The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m/s and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Answer:

Given: Mass m = 0.5 kg At O: PE = 30 J, velocity v = 0 m/s Total mechanical energy E = PE + KE = 30 + 0 = 30 J (constant) At points P, Q, R: Velocity v = sqrt[2(E - PE)/m] Calculate velocity at each point by substituting PE values from graph: For example, if PE at P = 10 J, v_P = sqrt[2(30 - 10)/0.5] = sqrt[2 × 20 / 0.5] = sqrt[80] ≈ 8.94 m/s Similarly calculate for Q and R using their PE values from the graph.

Explanation:

Total energy conserved. KE = Total energy - PE. Use KE to find velocity at each point.

MediumNCERT