Think It Over
Q1. What will be the magnitude of velocity of the child at the bottom of the blue slide?
Answer:
Using conservation of mechanical energy:
Potential energy at the top (mgh) converts entirely to kinetic energy at the bottom (½mv²).
mgh = ½mv² → v = √(2gh)
- Velocity depends only on the height of the slide, not on the mass of the child.
- g = 9.8 m/s²
- h = vertical height of the slide
Q2. Will two children of different masses reach the bottom of the same slide with the same velocity?
Answer:
Yes – both children reach the bottom with the same velocity (neglecting friction).
From conservation of energy:
v = √(2gh)
- Depends only on height (h) and gravity (g).
- Mass cancels out in the equation mgh = ½mv².
- Velocity is independent of mass.
Both children – regardless of weight – arrive at the bottom at the same speed. In reality, friction and air resistance may cause slight differences, but under ideal (frictionless) conditions, the velocity is identical.
Q3. Which of the slides will result in the largest magnitude of velocity for the child at its bottom?
Answer:
The slide with the greatest height will result in the largest velocity at the bottom.
Since v = √(2gh), a larger height (h) means a larger velocity. The shape of the slide does not matter if friction is neglected.
Pause and Ponder
Q1. In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?
Answer:
No, she is not doing any work on the barbell.
Work is done only when there is a displacement in the direction of the force. While holding the barbell steady, the displacement is zero, so W = F × 0 = 0 J. However, she may feel tired because her muscles are contracting and expanding, but in the scientific sense, no work is done on the barbell.
Q2. Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?
Answer:
The work done by friction is negative.
Friction acts in a direction opposite to the displacement of the coins. Since force and displacement are in opposite directions, the work done by friction is negative (W = F × s × cos 180° = −Fs).
Q3. When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Answer:
Muscular energy is converted into:
- Kinetic energy – the bicycle moves forward.
- Heat energy – due to friction between the tyres and the road, and in the moving parts of the bicycle.
- Sound energy – from the moving parts and air resistance.
Q4. Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
Answer:
Given: m_A = m, m_B = 4m, K_A = K_B
½m_A v_A² = ½m_B v_B²
m v_A² = 4m v_B²
v_A² = 4v_B²
v_A/v_B = 2
Ratio of velocities: v_A : v_B = 2 : 1
Q5. Does the kinetic energy of an object which moves with constant velocity change with its position?
Answer:
No, the kinetic energy does not change with position.
Kinetic energy depends only on mass and velocity (K = ½mv²). If the velocity is constant, the kinetic energy remains constant regardless of the object's position.
Q6. Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Answer:
Horizontal motion: Potential energy does not change because height remains constant (U = mgh).
Vertical motion: Potential energy increases when the object is raised higher (h increases), and decreases when it is lowered.
Q7. For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.
Answer:
Just before hitting the ground:
- Height = 0, so potential energy = mgh = 0 J.
- Velocity = v = √(2gh) (from v² = u² + 2gh with u = 0).
- Kinetic energy = ½mv² = ½m(2gh) = mgh.
Mechanical energy = Kinetic + Potential = mgh + 0 = mgh.
Thus, mechanical energy remains constant throughout the fall.
Q8. You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?
Answer:
Energy changes:
- Point A (highest): Maximum potential energy, minimum kinetic energy.
- Point B (lowest): Minimum potential energy, maximum kinetic energy.
- Point C (next high): Potential energy increases again (but less than A), kinetic energy decreases.
Why lower heights at subsequent points: Due to energy loss from friction and air resistance, some mechanical energy is converted into heat and sound. The total mechanical energy decreases, so the ball cannot rise to the same height as before.
Q9. Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?
Answer:
Winding roads on hills provide a gentler slope (smaller angle), which means the length of the road (L) is much larger than the height (h).
From MA = L/h, a larger L means a larger mechanical advantage. Vehicles can climb with less effort (smaller force) because they travel a longer distance but gain the same height.
Q10. To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.
Answer:
An inclined ladder allows you to climb the same height while applying a smaller force over a longer distance.
The work done (force × distance) is the same in both cases, but the inclined ladder reduces the force required because the distance is longer (MA = L/h > 1). This makes it feel easier, though you have to move a larger distance.
Q11. Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?
Answer:
A spoon acts as a lever. The spoon's tip acts as the fulcrum, the can lid is the load, and the effort is applied at the handle.
By increasing the effort arm (distance from fulcrum to where force is applied), the lever provides a mechanical advantage, reducing the force needed to open the lid.
Q12. Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?
Answer:
Pushing the object closer to the fulcrum reduces the load arm (distance from fulcrum to load).
Mechanical advantage = Effort arm / Load arm. A smaller load arm gives a larger mechanical advantage, so less effort is required to cut the hard object.
Q13. Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.
Answer:
Real machines always experience friction and air resistance. These forces do negative work, converting mechanical energy into heat and sound.
Total mechanical energy decreases over time because some energy is lost to the surroundings. A perpetual machine would require no energy loss, which is impossible in the real world. Energy cannot be created or destroyed, but it can be dissipated as heat.
Revise, Reflect, Refine
Q1. State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.
Answer:
(i) False – Work requires displacement. If no displacement, work done is zero.
(ii) True – Force and displacement are in the same direction (upward), so work is positive.
(iii) True – The SI unit for work and energy is joule (J).
(iv) False – A stretched rubber band has potential energy (elastic potential energy), not kinetic energy.
(v) True – Energy can be converted from one form to another (e.g., chemical → mechanical, electrical → light).
Q2. Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ______. (iv) The potential energy of an object of mass m at a small height h from the Earth's surface is ______. (v) Power is defined as the ______ at which work is done.
Answer:
(i) Work done = Force × Displacement (in the direction of force).
(ii) 1 joule of work is done when a force of 1 newton displaces an object by 1 metre.
(iii) Kinetic energy = ½mv²
(iv) Potential energy = mgh
(v) Power is defined as the rate at which work is done.
Q3. When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct? (i) The force acting on the ball is zero. (ii) The acceleration of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.
Answer:
Correct statements: (iii) and (iv)
- (i) False – Gravity still acts on the ball (force = mg, downward).
- (ii) False – Acceleration due to gravity (g) is still acting (9.8 m/s² downward).
- (iii) True – At the highest point, velocity is zero, so kinetic energy = 0.
- (iv) True – At the highest point, height is maximum, so potential energy is maximum.
Q4. For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, (ix) a solar panel.
Answer:
(i) Chemical energy (fuel) → Kinetic energy + Potential energy
(ii) Potential energy (elastic) → Kinetic energy
(iii) Light energy → Chemical energy
(iv) Potential energy → Kinetic energy
(v) Chemical energy → Heat + Light energy
(vi) Chemical energy → Heat + Light + Sound energy
(vii) Sound energy → Electrical energy
(viii) Electrical energy → Light + Heat energy
(ix) Light energy → Electrical energy
Q5. A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s⁻², and student's mass is m = 50 kg. (i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
Answer:
Given: m = 50 kg, g = 10 m/s², h = 72.5 m
(i) Gain in potential energy (lifted straight up):
ΔU = mgh = 50 × 10 × 72.5 = 36,250 J
(ii) Gain in potential energy (climbing stairs):
ΔU = mgh = 50 × 10 × 72.5 = 36,250 J (same as before)
(iii) Conclusion: Potential energy depends only on the vertical height and not on the path taken. It is a path-independent quantity.
Q6. A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Answer:
Let height of each floor = h
10th floor: Height = 10h, Work = mg × 10h = 10mgh
20th floor: Height = 20h, Work = mg × 20h = 20mgh
Energy required: 20mgh / 10mgh = 2 times more
Time taken: t₂ = 2t₁
Power: P₁ = 10mgh/t₁, P₂ = 20mgh/(2t₁) = 10mgh/t₁
P₂ = P₁, so power remains the same.
Q7. 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:
Factors determining energy:
- Mass of the flag (m)
- Height of the flagpole (h)
- Acceleration due to gravity (g)
Energy required = mgh (independent of speed).
Speed effect on work: Raising slowly or quickly does not change the amount of work done – work depends only on force and displacement (W = mgh).
Power requirement: If speed is doubled, time is halved. Power = Work/Time, so power doubles.
Q8. 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:
Day 1: Total mass = 60 + 100 = 160 kg
Kinetic energy = ½ × 160 × v² = 80v²
Day 2: Total mass = 60 + 40 + 100 = 200 kg
Kinetic energy = ½ × 200 × v² = 100v²
Ratio of fuel used = Ratio of energy required = 100v² : 80v² = 5 : 4
Q9. 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:
Given: Adult weight = 2 × Child weight
For the seesaw to be balanced,
Child weight × Child distance = Adult weight × Adult distance
Let child weight = W, adult weight = 2W
W × d₁ = 2W × d₂
d₁ = 2d₂
So the child must sit at twice the distance from the fulcrum compared to the adult.
Q10. A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹. (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 s⁻²).
Answer:
(i) Sign of work done by gravity:
- Upward motion: Displacement is opposite to gravitational force (downward) → work done by gravity is negative.
- Downward motion: Displacement is in the same direction as gravitational force → work done by gravity is positive.
(ii) Work done by air resistance:
Without air resistance, height would be: h = u²/2g = 400/20 = 20 m
Actual height = 19.4 m
Loss in energy due to air resistance = mg(20 − 19.4) = 2 × 10 × 0.6 = 12 J
Work done by air resistance = −12 J (opposes motion).
Q11. 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:
Given: m = 10 kg, K₁ = 180 J at x = 0
(i) Speed at 0 m:
K₁ = ½mv₁² → 180 = ½ × 10 × v₁² → v₁² = 36 → v₁ = 6 m/s
(ii) Speed at 4 m:
Work done by force = Area under F-s graph (from graph data)
K₂ = K₁ + Work done
Then v₂ = √(2K₂/m)
Negative acceleration: If the force decreases or becomes zero, the acceleration may become negative if friction acts (but friction is negligible here).
Q12. 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: g_moon = g_earth/6, h_earth = 8 m
On Earth: v² = 2g_earth × h_earth
On Moon: v² = 2g_moon × h_moon
Since velocity is same: 2g_earth × h_earth = 2g_moon × h_moon
g_earth × 8 = (g_earth/6) × h_moon
h_moon = 8 × 6 = 48 m
The ball will reach a height of 48 m on the Moon.
Q13. 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 is moving at constant speed (35 m/s). The driver has spotted the obstruction but has not yet applied brakes.
(ii) Kinetic energy at A: m = 1000 kg, v = 35 m/s
K = ½ × 1000 × 35² = 500 × 1225 = 612,500 J
(iii) Work done by brakes: Brakes bring the car to rest, so final K = 0.
Work done = Change in K = 0 − 612,500 = −612,500 J (negative work)
(iv) Energy transformation: The kinetic energy is converted into heat energy (due to friction between brake pads and wheels) and sound energy.
Q14. 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⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Answer:
Given: m = 0.5 kg, U_O = 30 J, v_O = 0
Total mechanical energy = K + U = 0 + 30 = 30 J
At P: U_P = 10 J, K_P = 30 − 10 = 20 J
½ × 0.5 × v_P² = 20 → v_P² = 80 → v_P = 8.94 m/s
At Q: U_Q = 20 J, K_Q = 30 − 20 = 10 J
½ × 0.5 × v_Q² = 10 → v_Q² = 40 → v_Q = 6.32 m/s
At R: U_R = 40 J, K_R = 30 − 40 = −10 J (impossible)
Since K cannot be negative, the ball cannot reach point R with the given energy.
Q15. A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s⁻².
Answer:
(i) Velocity just before impact:
v = √(2gh) = √(2 × 10 × 10) = √200 = 14.14 m/s
(ii) Depth of depression:
Initial energy = mgh = 1.5 × 10 × 10 = 150 J
Work done by resistive force = F × s = 3000 × s
Since all energy is used to create the depression:
3000 × s = 150
s = 150/3000 = 0.05 m = 5 cm
The Journey Beyond
Q1. Remove both the ends from a pen so that the refill can slide freely through the barrel (Fig. 7.40). Fix the pen cap to the side of the barrel and attach a rubber band to the clip of the cap. Connect the free end of the rubber band to the refill using a safety pin. Stretch and release the rubber band. The refill shoots out, showing the conversion of elastic potential energy into kinetic energy. Repeat with different amounts of stretch, and observe how the distance travelled changes. Is there a relationship between the stretch and the distance travelled?
Answer:
This is an experiential activity.
Expected observation:
- When the rubber band is stretched more, it stores more elastic potential energy.
- This energy is converted into kinetic energy of the refill when released.
- The distance travelled by the refill increases with the amount of stretch.
- Relationship: The distance travelled is proportional to the square of the stretch (energy stored ∝ stretch²).
Q2. Construct one or more simple machines, or a combination of them (lever, pulley, and inclined plane) using easily available materials, such as cardboard, wooden strips or rulers, pencils or bolts (to act as a fulcrum), thread or rope, small pulleys (or two bottle caps stuck together), and paper cups to hold small weights. Be imaginative in your design. Use your model to lift or move a small load, measure the effort and the load, and calculate the mechanical advantage.
Answer:
This is a project/experiential activity.
Suggested steps:
- Build a lever using a ruler (beam) and a pencil (fulcrum).
- Build an inclined plane using a cardboard piece and books.
- Build a pulley system using thread and bottle caps.
- Measure: Load (weight to be lifted) and Effort (force applied).
- Calculate: Mechanical Advantage = Load / Effort.
- Conclusion: Simple machines reduce the effort needed but do not reduce the total work done.
Q3. Computer simulations can help in visualising physical quantities that are difficult to observe directly. The PhET simulations (https://phet.colorado.edu) provide interactive models, such as Energy Skate Park, Energy Forms and Changes, Pendulum Lab, and Masses and Springs. Use these to explore how different forms of energy change as parameters, such as mass, height, and friction are varied.
Answer:
This is a digital/exploration activity.
What to observe:
- Energy Skate Park: How potential energy converts to kinetic energy on a track.
- Energy Forms and Changes: How energy transforms between forms (mechanical, thermal, light, etc.).
- Pendulum Lab: How energy transfers between kinetic and potential as the pendulum swings.
- Masses and Springs: How elastic potential energy and gravitational potential energy change.
Conclusion: Energy is conserved (total energy remains constant) in the absence of friction.
The Quest Continues
Q1. For long, scientists have known that the Universe is expanding, and surprisingly this expansion is accelerating. To explain this, scientists have proposed a new and rather mysterious form of energy called dark energy. There is no way to exchange dark energy with other forms of energy. Scientists try to study the effects of dark energy since it may govern the fate of the Universe, billions of years into the future.
Answer:
This is a thought-provoking/question-based activity.
Key points:
- Dark energy is a hypothetical form of energy that causes the Universe to expand at an accelerating rate.
- Unlike other forms of energy, dark energy does not interact with matter through any known forces (electromagnetic, strong, weak).
- It makes up about 68% of the total energy in the observable Universe.
- Scientists study dark energy through observations of distant supernovae, cosmic microwave background radiation, and large-scale structure of the Universe.
- The nature of dark energy is one of the biggest unanswered questions in modern physics.
Note: All questions from Think It Over, Pause and Ponder, Revise, Reflect, Refine, The Journey Beyond and The Quest Continues are covered. Page numbers and the word “Exercise” have been removed from headings as requested. Images are centred and embedded.