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Work, Energy, and Power - Work Done by a Variable Force

Grade 11CBSEPhysics

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

🔑Concepts

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A force is said to be variable if its magnitude or direction (or both) changes during the displacement of a body. Common examples include the spring force F=−kxF = -kx or gravitational force F=GMmr2F = \frac{GMm}{r^2}.

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If a variable force F(x)F(x) acts on a particle along the xx-axis, the work done dWdW for an infinitesimal displacement dxdx is given by dW=F(x)dxdW = F(x) dx.

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To find the total work done by a variable force from an initial position xix_i to a final position xfx_f, we calculate the definite integral of the force function over the displacement interval.

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Graphically, the work done by a variable force is equal to the area under the Force-Displacement (FF vs. xx) curve between the limits xix_i and xfx_f.

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For a spring following Hooke's Law, the restoring force is F=−kxF = -kx. The work done by an external agent to stretch a spring from x=0x=0 to x=xx=x is W=12kx2W = \frac{1}{2}kx^2.

📐Formulae

W=∫xixfF(x)dxW = \int_{x_i}^{x_f} F(x) dx

W=lim⁡Δx→0∑xixfF(x)ΔxW = \lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x) \Delta x

W=∫x1x2(Fxdx+Fydy+Fzdz)W = \int_{x_1}^{x_2} (F_x dx + F_y dy + F_z dz)

Wspring=∫0xkxdx=12kx2W_{\text{spring}} = \int_{0}^{x} kx dx = \frac{1}{2}kx^2

💡Examples

Problem 1:

A force F=(3x2+2x+5) NF = (3x^2 + 2x + 5) \text{ N} acts on a particle. Calculate the work done by this force in displacing the particle from x=0x = 0 to x=2 mx = 2 \text{ m}.

Solution:

The work done is given by: W=∫02(3x2+2x+5)dxW = \int_{0}^{2} (3x^2 + 2x + 5) dx Integrating each term: W=[3x33+2x22+5x]02W = \left[ \frac{3x^3}{3} + \frac{2x^2}{2} + 5x \right]_{0}^{2} W=[x3+x2+5x]02W = \left[ x^3 + x^2 + 5x \right]_{0}^{2} Substituting the upper and lower limits: W=(23+22+5(2))−(03+02+5(0))W = (2^3 + 2^2 + 5(2)) - (0^3 + 0^2 + 5(0)) W=(8+4+10)−0W = (8 + 4 + 10) - 0 Calculation of the sum: 84+1022\begin{array}{r} 8 \\ 4 \\ + 10 \\ \hline 22 \end{array} W=22 JW = 22 \text{ J}

Explanation:

We apply the definite integral of the force function F(x)F(x) over the given limits of displacement. The units are Joules (JJ) as per SI standards.

Problem 2:

A spring with spring constant k=500 N/mk = 500 \text{ N/m} is compressed by 0.1 m0.1 \text{ m}. Calculate the work done by the spring force.

Solution:

The work done by the spring force F=−kxF = -kx is: Ws=∫0x−kxdx=−12kx2W_s = \int_{0}^{x} -kx dx = -\frac{1}{2}kx^2 Substituting the values k=500k = 500 and x=0.1x = 0.1: Ws=−12×500×(0.1)2W_s = -\frac{1}{2} \times 500 \times (0.1)^2 Ws=−250×0.01W_s = -250 \times 0.01 Ws=−2.5 JW_s = -2.5 \text{ J}

Explanation:

The work done by the spring force is negative because the restoring force is opposite to the direction of displacement (compression).

Work Done by a Variable Force Class 11 Notes & Examples