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Work, Energy, and Simple Machines

Here is the revised, clean version of the “Revise, Reflect, Refine” solutions using clean, standard mathematical exponents (superscripts like x2, v2) so it looks polished and readable for your students while remaining easy to copy:

1. State whether True or False.

• (i) False — Work is only done when a force displaces an object in the direction of the force. If the object does not move, no work is done (W=F0=0).
• (ii) True — Lifting a bucket vertically upward applies an upward force, and the displacement is also in the upward direction, resulting in positive work done.
• (iii) True — The SI unit for both work and energy is the joule (J).
• (iv) False — A motionless stretched rubber band possesses potential energy (specifically elastic potential energy) due to its deformation, not kinetic energy.
• (v) True — Energy is conserved and can freely transform from one form to another (e.g., potential energy to kinetic energy).

2. Fill in the blanks.

• (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 in the direction of the force.
• (iii) The expression for kinetic energy of a body of mass m and velocity v is 12mv2.
• (iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.
• (v) Power is defined as the rate at which work is done.

3. When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?

• (iii) Its kinetic energy is zero. (Correct, because the ball momentarily comes to rest at its highest point, meaning v=0).
• (iv) Its potential energy is maximum. (Correct, because it is at its maximum height). (Note: Statements (i) and (ii) are incorrect because gravity still acts downward on the ball with an acceleration of g, so neither the net force nor the acceleration is zero).

4. Identify the energy transformation for each situation:

• (i) A truck moving uphill: Chemical energy of fuel Kinetic energy + Potential energy.
• (ii) Unwinding of a watch spring: Elastic potential energy Kinetic energy (mechanical energy of the gears/hands).
• (iii) Photosynthesis in green leaves: Light (solar) energy Chemical energy.
• (iv) Water flowing from a dam: Gravitational potential energy Kinetic energy.
• (v) Burning of a matchstick: Chemical energy Thermal (heat) energy + Light energy.
• (vi) Explosion of a firecracker: Chemical energy Sound energy + Light energy + Thermal energy + Kinetic energy.
• (vii) Speaking into a microphone: Sound (acoustic) energy Electrical energy.
• (viii) A glowing electric bulb: Electrical energy Light energy + Thermal energy.
• (ix) A solar panel: Light (solar) energy Electrical energy.

5. Student in an elevator vs. climbing stairs (h=72.5m, g=10ms-2, mass m=50kg):

• (i) Gain in potential energy (elevator):
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• U=mgh=50kg10ms-272.5m=36,250J
• (ii) Gain in potential energy (climbing stairs):
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• U=mgh=50kg10ms-272.5m=36,250J
• (iii) Conclusion: Gravitational potential energy depends only on the initial and final vertical positions and is independent of the path taken.

6. Crane lifting a mass:

• Energy Required: The 20th floor is twice as high as the 10th floor (h20=2h10). Since potential energy is U=mgh, doubling the height doubles the potential energy. Thus, twice as much energy is required.
• Power Required: Power is energy divided by time (P=Et). The energy is doubled (2E), but the time taken is also doubled (2t):
• 
  
• Pnew=2E2t=Et
• Therefore, the power required remains the same.

7. Raising a flag using a pulley:

• Factors determining energy: The energy required depends on the mass of the flag (m), acceleration due to gravity (g), and the height of the flagpole (h) (E=mgh).
• Does speed affect work done? No. Work done depends only on force and displacement (W=Fh). Raising it slowly or quickly does not change the total work done.
• Power requirement if speed is doubled: Power is the rate of doing work (P=Wt=Fv). If the speed (v) is doubled while the force remains constant, the power requirement is doubled.

8. Scooter fuel consumption ratio:

• Let the mass of the man be m1=60kg and the scooter be ms=100kg.
• Day 1 total mass: M1=60+100=160kg.
• Day 2 total mass (with son, 40kg): M2=160+40=200kg.
• Since both reach the same final velocity v in the same time interval, the energy transferred from the fuel goes entirely into the kinetic energy of the system (KE=12Mv2).
• Ratio of fuel (energy) used:
• 
  
• E1E2=12M1v212M2v2=M1M2=160200=45
• Thus, the ratio of fuel used is 4:5.

9. Seesaw with a child and an adult:

• To balance a seesaw, the product of force (weight) and distance from the fulcrum must be equal on both sides (Law of Moments: F1d1=F2d2).
• Since the adult weighs twice as much as the child (Fadult=2Fchild), the child must sit twice as far from the fulcrum as the adult.
• Example sketch layout: If the adult sits at a distance of 1meter from the fulcrum on one side, the child must sit at a distance of 2meters on the opposite side.

10. Ball of mass 2kg thrown up (v=20ms-1, height reached h=19.4m):

• (i) Sign of work done by gravity:
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  • During upward motion: Gravity acts downward while displacement is upward Negative work (=180).
  • During downward motion: Gravity acts downward and displacement is downward Positive work (=0).
• (ii) Work done by air resistance:
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  • Initial Kinetic Energy (KEi) = 12mv2=12(2)(202)=400J.
  • Final Potential Energy at peak (PEf) = mgh=(2)(10)(19.4)=388J.
  • Total initial energy = 400J. Total final potential energy = 388J.
  • Energy lost against air resistance = 400J-388J=12J.
  • Therefore, the work done by air resistance is -12J (or 12J opposing the motion).

11. Block on horizontal floor (mass m=10.0kg, initial KE=180J at 0m):

• (i) Speed at 0m:
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• KE = (1/2) * m * v² -> 180 = (1/2) * 10 * v² -> 5 * v² = 180 -> v² = 36 -> v = 6 m/s
• (ii) Speed at 4m:
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  • Work done equals the area under the Force-Displacement graph (Fig. 7.37).
  • The graph is a trapezium with parallel sides of length 4m (from 0 to 4) and 2m (from 1 to 3), and a height of 50N.
  • WorkDone=Area=12(sumofparallelsides)height=12(4+2)50=150J.
  • Final KE=InitialKE+WorkDone=180J+150J=330J.
  • (1/2) * m * v_f² = 330 -> (1/2) * 10 * v_f² = 330 -> 5 * v_f² = 330 -> v_f² = 66 -> v_f = sqrt(66) ≈ 8.12 m s^-1

12. Ball thrown on the Moon:

• On Earth, height hE=8m. Initial kinetic energy given to the ball is the same in both cases: KE=mgEhE.
• On the Moon, gravity is gM=16gE.
• Equating initial energy to potential energy at peak:
• 
  
• m * g_E * h_E = m * g_M * h_M -> m * g_E * (8) = m * (1/6 * g_E) * h_M
• 8 = (1/6) * h_M -> h_M = 8 * 6 = 48 m

13. Car motion analysis (m=1000kg, Fig. 7.38):

• (i) Motion between A and B: The speed-time graph is a horizontal straight line at 35ms-1 from t=0 to 1s. This means the car moves at a constant speed of 35ms-1 (driver’s reaction time).
• (ii) Kinetic energy at A:
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• KE=12mv2=12(1000)(352)=5001225=612,500J
• (iii) Work done by brakes between B and C:
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  • At point C, the car comes to a complete stop (v=0), so final KE=0.
  • By the work-energy theorem, work done by brakes = Change in kinetic energy = 0-612,500J=-612,500J.
• (iv) Transformation of kinetic energy: The kinetic energy transforms into heat (thermal) energy and sound energy due to friction in the braking mechanism and between tires and the road surface.

14. Ball on a frictionless track (m=0.5kg, Fig. 7.39):

• At point O, velocity vO=0ms-1, so KEO=0.
• Given Potential Energy at O is 30J, the Total Mechanical Energy (E) is constant:
• 
  
• E=PE+KE=30J+0=30J
• Using E=PE+12mv2:
• 
  
  • At point P (PEP=20J):
  • 
    

KE_P = 30 – 20 = 10 J -> (1/2) * 0.5 * v_P² = 10 -> 0.25 * v_P² = 10 -> v_P² = 40 -> v_P = sqrt(40) ≈ 6.32 m s^-1

At point Q (PE_Q = 30 J):

• KE_Q = 30 – 30 = 0 J -> v_Q = 0 m s^-1

• At point R (PER=40J): Since total energy is 30J and PE at R is 40J, the ball cannot reach point R without external energy.

15. Coconut falling into sand (m=1.5kg, height h=10m, resistive force F=3000N):

• (i) Velocity just before hitting the sand:
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• v² = u² + 2 * g * h = 0 + 2 * 10 * 10 = 200 -> v = sqrt(200) = 10 * sqrt(2) ≈ 14.14 m s^-1
• (ii) Depth of the depression (d):
• 
  
  • Total mechanical energy upon impact = mgh=1.5kg10ms-210m=150J.
  • Energy used up by resistive force:
  • WorkDone=Fd=150J
  • 3000 N * d = 150 J -> d = 150 / 3000 = 0.05 m (or 5 cm)