Work of the forces

Classical Mechanics Level 3

Two forces ( 6 i ^ + 2 j ^ − 3 k ^ ) N \left(6\hat{i} + 2\hat{j} - 3\hat{k}\right) \si{\newton} and ( 5 i ^ − 3 j ^ + 7 k ^ ) N \left(5\hat{i} - 3\hat{j} + 7\hat{k}\right) \si{\newton} act on an object that moves on a frictionless surface and, in doing so, displace it from ( 2 i ^ + 3 j ^ − 5 k ^ ) m \left(2\hat{i} + 3\hat{j} - 5\hat{k}\right) \si{\metre} to ( − 3 i ^ − 3 j ^ + 4 k ^ ) m \left(-3\hat{i} - 3\hat{j} + 4\hat{k}\right) \si{\metre} .

Find the magnitude of the work done on the object, W W in joules.


The answer is 13.00.

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Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Jul 5, 2016

The sum total of the forces acting on the body is ( ( 6 + 5 ) i ^ + ( 2 − 3 ) j ^ + ( − 3 + 7 ) k ^ ) N = ( 11 i ^ − j ^ + 4 k ^ ) N \left((6+5)\hat{i} + (2-3)\hat{j} + (-3+7)\hat{k}\right)N = \left(11\hat{i} - \hat{j} + 4\hat{k}\right)N .

The displacement caused is ( ( − 3 − 2 ) i ^ + ( − 3 − 3 ) j ^ + ( 4 − ( − 5 ) ) k ^ ) m = ( − 5 i ^ − 6 j ^ + 9 k ^ ) m \left((-3-2)\hat{i} + (-3-3)\hat{j} + (4-(-5))\hat{k}\right)m = \left(-5\hat{i} - 6\hat{j} + 9\hat{k}\right)m .

W ⃗ = F ⃗ ⋅ S ⃗ = ( ( 11 i ^ − j ^ + 4 k ^ ) ⋅ ( − 5 i ^ − 6 j ^ + 9 k ^ ) ) J = ( 11 × − 5 ) + ( − 1 × − 6 ) + ( 4 × 9 ) = − 55 + 6 + 36 = − 13 J ∴ ∣ W ∣ = 13 \begin{aligned} \vec{W} & = \vec{F} \cdot \vec{S}\\ \\ & = \left(\left(11\hat{i} - \hat{j} + 4\hat{k}\right) \cdot \left(-5\hat{i} - 6\hat{j} + 9\hat{k}\right)\right)J\\ \\ & = (11 × -5) + (-1 × -6) + (4 × 9)\\ \\ & = -55 + 6 + 36\\ \\ & = -13 J\\ \\ \therefore |W| & = \color{#3D99F6}{\boxed{13}} \end{aligned}

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