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Work, Energy, and Power - Work-Energy Theorem

Grade 11CBSEPhysics

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

🔑Concepts

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The Work-Energy Theorem states that the net work done by the sum of all forces (resultant force) acting on a particle is equal to the change in the kinetic energy of the particle.

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The theorem is applicable to both constant and variable forces.

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Kinetic Energy (KK) is the energy possessed by an object due to its motion, defined as K=12mv2K = \frac{1}{2}mv^2.

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If the net work done WnetW_{net} is positive, the kinetic energy increases (Kf>KiK_f > K_i), meaning the particle speeds up.

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If the net work done WnetW_{net} is negative, the kinetic energy decreases (Kf<KiK_f < K_i), meaning the particle slows down.

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For a variable force, the work done is the integral of the force over displacement: W=∫xixfF(x)dxW = \int_{x_i}^{x_f} F(x) dx, which by the theorem equals ΔK\Delta K.

📐Formulae

Wnet=ΔKW_{net} = \Delta K

Wnet=Kf−KiW_{net} = K_f - K_i

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

Wnet=12mvf2−12mvi2W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

∫xixfFnetdx=12mvf2−12mvi2\int_{x_i}^{x_f} F_{net} dx = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

💡Examples

Problem 1:

A block of mass m=2 kgm = 2\text{ kg} is moving on a frictionless horizontal surface with an initial velocity of u=5 m/su = 5\text{ m/s}. A constant force is applied on the block in the direction of motion, increasing its velocity to v=10 m/sv = 10\text{ m/s}. Calculate the work done by the force.

Solution:

W=ΔK=12m(v2−u2)W = \Delta K = \frac{1}{2}m(v^2 - u^2) W=12×2×(102−52)W = \frac{1}{2} \times 2 \times (10^2 - 5^2) W=1×(100−25)W = 1 \times (100 - 25) W=75 JW = 75\text{ J}

Explanation:

According to the Work-Energy Theorem, the work done by the external force is equal to the change in kinetic energy. Since the mass and both velocities are known, we calculate the difference between final and initial kinetic energies.

Problem 2:

A bullet of mass 20 g20\text{ g} (0.02 kg0.02\text{ kg}) moving with a speed of 100 m/s100\text{ m/s} enters a heavy wooden block and is stopped after a distance of 50 cm50\text{ cm} (0.5 m0.5\text{ m}). What is the average resistive force exerted by the block on the bullet?

Solution:

Wnet=ΔKW_{net} = \Delta K Since the bullet stops, vf=0v_f = 0. F⋅d⋅cos⁡(180∘)=12mvf2−12mvi2F \cdot d \cdot \cos(180^\circ) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 −F⋅0.5=0−12(0.02)(100)2-F \cdot 0.5 = 0 - \frac{1}{2}(0.02)(100)^2 −0.5F=−0.01×10000-0.5F = -0.01 \times 10000 −0.5F=−100-0.5F = -100 F=1000.5=200 NF = \frac{100}{0.5} = 200\text{ N}

Explanation:

The work done by the resistive force is negative because the force is opposite to the displacement. By setting the work done (F⋅dF \cdot d) equal to the change in kinetic energy, we can solve for the unknown resistive force FF.