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Work, Energy and Power - Work done by Constant and Variable Forces

Grade 11ICSEPhysics

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

🔑Concepts

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Work is defined as the dot product of the force vector F⃗\vec{F} and the displacement vector s⃗\vec{s}: W=F⃗⋅s⃗=Fscos⁡θW = \vec{F} \cdot \vec{s} = Fs \cos \theta, where θ\theta is the angle between the force and displacement.

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Work is a scalar quantity. Its SI unit is the Joule (JJ) and its CGS unit is the erg. Note: 1 J=107 erg1 \, J = 10^7 \, erg.

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Dimensional formula of work is [M1L2T−2][M^1L^2T^{-2}].

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Work can be positive, negative, or zero depending on θ\theta: Positive if 0∘≤θ<90∘0^\circ \le \theta < 90^\circ, Negative if 90∘<θ≤180∘90^\circ < \theta \le 180^\circ, and Zero if θ=90∘\theta = 90^\circ or if displacement is zero.

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Work done by a Variable Force: When the force changes with position, work is calculated by integrating the force over the path: W=∫x1x2F(x) dxW = \int_{x_1}^{x_2} F(x) \, dx.

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Graphical Interpretation: The area under a Force-Displacement (FF-ss) graph represents the work done. For variable forces, the area is calculated using the integral under the curve.

📐Formulae

W=Fscos⁡θW = Fs \cos \theta

W=F⃗⋅s⃗=Fxx+Fyy+FzzW = \vec{F} \cdot \vec{s} = F_x x + F_y y + F_z z

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

W=∫x1x2Fx dx+∫y1y2Fy dy+∫z1z2Fz dzW = \int_{x_1}^{x_2} F_x \, dx + \int_{y_1}^{y_2} F_y \, dy + \int_{z_1}^{z_2} F_z \, dz

1 Joule=107 ergs1 \text{ Joule} = 10^7 \text{ ergs}

💡Examples

Problem 1:

A force of 10 N10 \, N acts on a block at an angle of 60∘60^\circ to the horizontal. If the block is displaced 5 m5 \, m along the horizontal floor, calculate the work done.

Solution:

W=Fscos⁡θW = Fs \cos \theta W=10×5×cos⁡60∘W = 10 \times 5 \times \cos 60^\circ W=50×12=25 JW = 50 \times \frac{1}{2} = 25 \, J.

Explanation:

Since the force is constant and the displacement is along a straight line, we use the scalar product formula involving the cosine of the angle between the force and displacement vectors.

Problem 2:

A particle moves from x=0x = 0 to x=3 mx = 3 \, m under the influence of a force F=(2x+3) NF = (2x + 3) \, N. Calculate the work done by this force.

Solution:

W=∫x1x2F dxW = \int_{x_1}^{x_2} F \, dx W=∫03(2x+3) dxW = \int_{0}^{3} (2x + 3) \, dx W=[x2+3x]03W = [x^2 + 3x]_{0}^{3} W=(32+3(3))−(02+3(0))W = (3^2 + 3(3)) - (0^2 + 3(0)) W=9+9=18 JW = 9 + 9 = 18 \, J.

Explanation:

Because the force is a function of position (xx), it is a variable force. The work done is the definite integral of the force function with respect to xx over the specified limits.

Problem 3:

Calculate the work done by a force F⃗=(3i^+4j^) N\vec{F} = (3\hat{i} + 4\hat{j}) \, N in displacing an object from position r⃗1=(2i^+1j^) m\vec{r}_1 = (2\hat{i} + 1\hat{j}) \, m to r⃗2=(5i^+5j^) m\vec{r}_2 = (5\hat{i} + 5\hat{j}) \, m.

Solution:

Displacement s⃗=r⃗2−r⃗1\vec{s} = \vec{r}_2 - \vec{r}_1 s⃗=(5−2)i^+(5−1)j^=3i^+4j^ m\vec{s} = (5-2)\hat{i} + (5-1)\hat{j} = 3\hat{i} + 4\hat{j} \, m. W=F⃗⋅s⃗=(3i^+4j^)⋅(3i^+4j^)W = \vec{F} \cdot \vec{s} = (3\hat{i} + 4\hat{j}) \cdot (3\hat{i} + 4\hat{j}) W=(3×3)+(4×4)=9+16=25 JW = (3 \times 3) + (4 \times 4) = 9 + 16 = 25 \, J.

Explanation:

In vector notation, work is the dot product of the constant force vector and the net displacement vector.

Work done by Constant and Variable Forces Revision - Class 11 Physics ICSE