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Work, Energy and Power - Work-Energy Theorem

Grade 11ICSEPhysics

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 resultant force acting on a body is equal to the change in its kinetic energy: Wnet=ΔKW_{net} = \Delta K.

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This theorem is applicable for both constant and variable forces. For a constant force, W=F⋅s=12mv2−12mu2W = F \cdot s = \frac{1}{2}mv^2 - \frac{1}{2}mu^2.

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For a variable force, the work done is represented by the integral: W=∫xixfF(x)dxW = \int_{x_i}^{x_f} F(x) dx.

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If the net work done is positive (W>0W > 0), the kinetic energy increases (Kf>KiK_f > K_i). If the net work done is negative (W<0W < 0), the kinetic energy decreases (Kf<KiK_f < K_i).

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The theorem accounts for work done by all types of forces, including conservative, non-conservative, internal, and external forces.

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Kinetic Energy (KK) can also be expressed in terms of linear momentum (pp) as K=p22mK = \frac{p^2}{2m}, where mm is the mass of the object.

📐Formulae

Wnet=ΔK=Kf−KiW_{net} = \Delta K = K_f - K_i

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

W=∫vivfmv dv=12mvf2−12mvi2W = \int_{v_i}^{v_f} m v \, dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

K=p22mK = \frac{p^2}{2m}

W=F⃗⋅d⃗=Fdcos⁡θW = \vec{F} \cdot \vec{d} = Fd \cos \theta

💡Examples

Problem 1:

A bullet of mass 20 g20 \text{ g} 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}. Calculate the average resistive force exerted by the block using the Work-Energy Theorem.

Solution:

Given: m=20 g=0.02 kgm = 20 \text{ g} = 0.02 \text{ kg}, u=100 m/su = 100 \text{ m/s}, v=0 m/sv = 0 \text{ m/s}, s=50 cm=0.5 ms = 50 \text{ cm} = 0.5 \text{ m}.\According to the Work-Energy Theorem: W=ΔKW = \Delta K\−F⋅s=12m(v2−u2)-F \cdot s = \frac{1}{2}m(v^2 - u^2)\−F×0.5=12×0.02×(02−1002)-F \times 0.5 = \frac{1}{2} \times 0.02 \times (0^2 - 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 opposes the motion. By equating this work to the change in kinetic energy (which is negative since the bullet stops), we find the magnitude of the average force.

Problem 2:

A body of mass 5 kg5 \text{ kg} is acted upon by a variable force F=(3x2+2x) NF = (3x^2 + 2x) \text{ N}. Calculate the change in kinetic energy of the body as it moves from x=0 mx = 0 \text{ m} to x=2 mx = 2 \text{ m}.

Solution:

From the Work-Energy Theorem, ΔK=W=∫xixfF(x)dx\Delta K = W = \int_{x_i}^{x_f} F(x) dx.\W=∫02(3x2+2x)dxW = \int_{0}^{2} (3x^2 + 2x) dx\W=[x3+x2]02W = [x^3 + x^2]_{0}^{2}\W=(23+22)−(03+02)W = (2^3 + 2^2) - (0^3 + 0^2)\W=8+4=12 JW = 8 + 4 = 12 \text{ J}.\Therefore, ΔK=12 J\Delta K = 12 \text{ J}.

Explanation:

Since the force is a function of position xx, we integrate the force with respect to displacement to find the total work done, which equals the change in kinetic energy.

Work-Energy Theorem - Revision Notes & Key Formulas | ICSE Class 11 Physics