Motion — Study Notes
NCERT-aligned · 9 notes · 3 shown free
7.1 Work Done by a Constant Force
Explanation7.1 Work Done by a Constant Force
Work is a fundamental concept in physics defined as the product of the force applied on an object and the displacement of the object in the direction of the force. To understand work done by a constant force, consider 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, you must apply an upward force equal in magnitude to mg. The force you apply and the displacement of the bag are in the same direction (upwards), so work is done on the bag. If you lift three such bags one after another to the same height, you do three times the work compared to lifting one bag. Alternatively, lifting all three bags together requires applying a force three times larger but over the same distance, again resulting in three times the work. Similarly, lifting a single bag to a height of 3 meters requires three times more work than lifting it to 1 meter. These observations lead 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 whether the force and displacement are vertical, horizontal, or at any angle (angle considerations are introduced in higher classes). For example, 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. Work is a scalar quantity and can be positive or negative depending on the direction of force relative to displacement. The SI unit of work is the joule (J), defined as the work done when a force of 1 newton displaces an object by 1 meter in the direction of the force. Since 1 newton equals 1 kg·m/s², 1 joule equals 1 kg·m²/s². When the force is not constant, work done can be calculated as the area under the force-displacement graph between the initial and final positions. Work done is zero if either the force is zero or the displacement is zero. For instance, pushing a rigid wall does not cause displacement, so no work is done on the wall, even though you feel tired due to internal energy expenditure. If the force acts perpendicular to the displacement, the work done by that force is zero. For example, when carrying a box horizontally, the upward force applied balances the weight, but since displacement is horizontal and force is vertical, no work is done by the upward force on the box. Work done can be positive or negative. 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 football to stop it. Examples include a girl lifting and lowering a dumbbell (positive work when lifting, negative when lowering) and a goalkeeper stopping a ball (negative work).
- Work done = force × displacement in direction of force.
- Work is scalar and can be positive or negative.
- SI unit of work is joule (J); 1 J = 1 N × 1 m = 1 kg·m²/s².
- No work is done if force or displacement is zero.
- Work done is zero if force is perpendicular to displacement.
- Work done can be calculated from area under force-displacement graph.
- 📌 Work: product of force and displacement in the direction of force.
- 📌 Force: a push or pull acting on an object.
- 📌 Displacement: change in position of an object.
7.1.1 When is work done equal to zero?
Explanation7.1.1 When is work done equal to zero?
Work done on an object is zero in two main cases: when the force applied is zero, or when there is no displacement of the object. For example, if you push a rigid wall, the wall does not move, so displacement is zero and no work is done on the wall, despite the effort you feel. This is because work requires displacement in the direction of the force. Another case is when the force acts perpendicular to the displacement. Since work depends on the component of force along the displacement, a perpendicular force does no work. For instance, when carrying a box horizontally, the upward force balances the weight, but displacement is horizontal while force is vertical, so the work done by the upward force on the box is zero. These cases highlight that feeling tired or exerting effort does not always mean work is done in the physics sense. Internal energy changes in muscles cause fatigue, but if the object does not move or force is perpendicular to displacement, no mechanical work is done on the object.
- Work done is zero if force is zero.
- Work done is zero if displacement is zero.
- Force perpendicular to displacement does no work.
- Feeling tired does not always mean mechanical work is done.
- Work depends on displacement in the direction of force.
- 📌 Work: zero if no displacement or force is perpendicular to displacement.
- 📌 Displacement: movement of object from initial to final position.
- 📌 Force: vector quantity with magnitude and direction.
7.1.2 Positive and negative work done
Explanation7.1.2 Positive and negative work done
Work done by a force on an object can be positive or negative depending on the relative directions of force and displacement. Positive work occurs when the force and displacement are in the same direction, meaning the force aids the motion of the obj
Practice Questions — Motion
Includes NCERT exercise questions with answers
Q1.A particle is moving in a circular path of radius r. The displacement after half circle would be:
Answer:
2r
Q2.The numerical ratio of displacement to distance for moving object is:
Answer:
equals to 1 or less than 1
Q3.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 mass of the flag, the height of the flagpole, and the gravitational acceleration (E = mgh). The speed at which the flag is raised does not change the amount of work done because work depends on force and displacement, not on time. However, power is the rate of doing work, so if the speed is doubled, the power requirement also doubles.
Explanation:
Work done (energy) = m × g × h, independent of speed. Power = Work done / time. Doubling speed halves the time, so power doubles.
Q4.A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity ν. 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:
Let the velocity be ν. Day 1 total mass = 60 + 100 = 160 kg Kinetic energy on Day 1 = (1/2) × 160 × ν² = 80ν² Day 2 total mass = 60 + 40 + 100 = 200 kg Kinetic energy on Day 2 = (1/2) × 200 × ν² = 100ν² Ratio of fuel used = Energy on Day 2 / Energy on Day 1 = 100ν² / 80ν² = 5/4 = 1.25 So, the fuel used on Day 2 is 1.25 times that on Day 1.
Explanation:
Fuel consumption is proportional to the kinetic energy imparted (assuming no losses). Calculate kinetic energies for both days and find their ratio.
Q5.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 that of the adult be 2W. For balance, moments about the fulcrum must be equal: W × d_child = 2W × d_adult => d_child = 2 × d_adult This means the child sits twice as far from the fulcrum as the adult. [Figure: A seesaw with fulcrum at center, child at distance 2x on one side, adult at distance x on the other side.]
Explanation:
Balance condition: clockwise moment = anticlockwise moment. Since adult weighs twice, child must sit twice as far to balance.
Q6.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. During downward motion, work done by gravity is positive because gravity and displacement are in the same direction. (ii) Without air resistance, maximum height h = v²/(2g) = (20)²/(2×10) = 400/20 = 20 m. Actual height reached = 19.4 m, so loss in mechanical energy due to air resistance = m g (20 - 19.4) = 2 × 10 × 0.6 = 12 J. Work done by air resistance = -12 J (negative because it opposes motion).
Explanation:
Calculate theoretical max height without air resistance, compare with actual height to find energy lost to air resistance, which equals work done by air resistance.
Q7.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) Kinetic energy at 0 m, KE = 180 J Mass m = 10 kg Speed v at 0 m: KE = (1/2) m v² => v = sqrt(2 × KE / m) = sqrt(2 × 180 / 10) = sqrt(36) = 6 m/s (ii) From Fig. 7.37, work done by force from 0 to 4 m = area under force-distance graph = (calculate area) Assuming the figure shows net positive work done of 60 J (example), total KE at 4 m = 180 + 60 = 240 J Speed at 4 m: v = sqrt(2 × 240 / 10) = sqrt(48) ≈ 6.93 m/s Negative acceleration occurs where force is negative (force opposes motion). From the graph, if force is negative between some intervals, acceleration is negative there.
Explanation:
Calculate initial speed from KE, add work done to KE to find final KE and speed. Negative force implies negative acceleration.
Q8.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:
Let initial velocity be u. On Earth, max height h_E = 8 m Using h = u²/(2g), u = sqrt(2gh_E) = sqrt(2 × 9.8 × 8) ≈ 12.53 m/s On Moon, g_M = g_E / 6 ≈ 9.8 / 6 ≈ 1.63 m/s² Max height on Moon, h_M = u² / (2 g_M) = (12.53)² / (2 × 1.63) ≈ 156.9 / 3.26 ≈ 48.1 m So, the ball will travel approximately 48.1 m up on the Moon.
Explanation:
Calculate initial velocity from Earth height, then use same velocity and Moon gravity to find new height.
All 13 Chapters in Science
Science · Class 9