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Work and Energy - Conservative and Non-Conservative Forces-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Conservative Forces: A force is said to be conservative if the work done by or against the force in moving a body from one point to another depends only on the initial and final positions of the body and not on the nature of the path followed. Examples include Gravitational force, Electrostatic force, and Elastic spring force.

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Path Independence: For conservative forces, the work done along any closed path is zero: ∮F⃗⋅dr⃗=0\oint \vec{F} \cdot d\vec{r} = 0. This implies that energy can be recovered completely.

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Potential Energy (UU): Potential energy is only defined for conservative forces. The change in potential energy is equal to the negative of the work done by the conservative force: ΔU=−Wc\Delta U = -W_{c}.

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Non-Conservative Forces: A force is non-conservative if the work done by or against it depends on the path taken between two points. Examples include Friction, Air Resistance, and Viscous force. Energy is usually dissipated as heat or sound.

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Total Mechanical Energy: The sum of kinetic energy (KK) and potential energy (UU). In the presence of only conservative forces, the total mechanical energy remains constant: Ki+Ui=Kf+UfK_i + U_i = K_f + U_f.

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Work-Energy Theorem (Advanced): The work done by all forces (conservative and non-conservative) equals the change in kinetic energy: Wc+Wnc=ΔKW_{c} + W_{nc} = \Delta K. This can be rewritten as Wnc=ΔK+ΔUW_{nc} = \Delta K + \Delta U.

📐Formulae

W=Fscos⁡θW = F s \cos \theta

K=12mv2K = \frac{1}{2}mv^2

U=mghU = mgh

Wnet=ΔKW_{net} = \Delta K

Wconservative=−ΔUW_{conservative} = -\Delta U

Etotal=K+U=constant (for conservative forces only)E_{total} = K + U = \text{constant (for conservative forces only)}

F=−dUdx (Relation between Force and Potential Energy)F = -\frac{dU}{dx} \text{ (Relation between Force and Potential Energy)}

💡Examples

Problem 1:

A ball of mass m=0.5 kgm = 0.5\text{ kg} is dropped from a height of 10 m10\text{ m}. It hits the ground and rebounds to a height of 8 m8\text{ m}. Calculate the work done by the non-conservative force (air resistance and energy lost during impact) during the entire process. (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

  1. Initial Mechanical Energy at height h1=10 mh_1 = 10\text{ m}: Ei=mgh1=0.5×10×10=50 JE_i = mgh_1 = 0.5 \times 10 \times 10 = 50\text{ J}
  2. Final Mechanical Energy at rebound height h2=8 mh_2 = 8\text{ m}: Ef=mgh2=0.5×10×8=40 JE_f = mgh_2 = 0.5 \times 10 \times 8 = 40\text{ J}
  3. Work done by non-conservative forces (WncW_{nc}): Wnc=Ef−EiW_{nc} = E_f - E_i Wnc=40−50=−10 JW_{nc} = 40 - 50 = -10\text{ J}

Explanation:

Since the ball did not reach the original height, energy was lost. This loss (10 J10\text{ J}) represents the negative work done by non-conservative forces like air friction and heat/sound during impact.

Problem 2:

Calculate the work done by friction when a block of mass 2 kg2\text{ kg} is pushed along a horizontal floor for 5 m5\text{ m} and then pushed back to its starting point. Assume the coefficient of friction μ=0.1\mu = 0.1 and g=10 m/s2g = 10\text{ m/s}^2.

Solution:

  1. Frictional force (ff): f=μmg=0.1×2×10=2 Nf = \mu mg = 0.1 \times 2 \times 10 = 2\text{ N}
  2. Work done during forward trip (s=5 ms = 5\text{ m}): W1=f×s×cos⁡(180∘)=2×5×(−1)=−10 JW_1 = f \times s \times \cos(180^{\circ}) = 2 \times 5 \times (-1) = -10\text{ J}
  3. Work done during return trip (s=5 ms = 5\text{ m}): W2=f×s×cos⁡(180∘)=2×5×(−1)=−10 JW_2 = f \times s \times \cos(180^{\circ}) = 2 \times 5 \times (-1) = -10\text{ J}
  4. Total work done by friction: Wtotal=W1+W2=−10+(−10)=−20 JW_{total} = W_1 + W_2 = -10 + (-10) = -20\text{ J}

Explanation:

Unlike gravity (a conservative force), the work done by friction in a round trip is not zero. Friction always opposes motion, so work is negative in both directions, making it path-dependent and non-conservative.