Work Energy Power

JEE Main level, numerical answer, medium

4 of 183

Question 4 of 183

A 11 kg block subjected to two simultaneous forces (2i^+3j^+4k^)(2\hat{i} + 3\hat{j} + 4\hat{k}) N and (3i^−j^−2k^)(3\hat{i} - \hat{j} - 2\hat{k}) N is moved a distance of 2525 m along (3i^−4j^)(3\hat{i} - 4\hat{j}) direction. The work done in this process is _____ J.

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Answer: 35

The net force acting on the block is the vector sum of the two forces:

F⃗=F⃗1+F⃗2=(2i^+3j^+4k^)+(3i^−j^−2k^)=5i^+2j^+2k^\vec{F} = \vec{F}_1 + \vec{F}_2 = (2\hat{i} + 3\hat{j} + 4\hat{k}) + (3\hat{i} - \hat{j} - 2\hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k} N

The block is moved a distance of 2525 m along the direction of the vector 3i^−4j^3\hat{i} - 4\hat{j}. The unit vector along this direction is:

n^=3i^−4j^32+(−4)2=3i^−4j^5\hat{n} = \dfrac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \dfrac{3\hat{i} - 4\hat{j}}{5}

The displacement vector s⃗\vec{s} is the magnitude multiplied by the unit vector:

s⃗=25n^=25(3i^−4j^5)=15i^−20j^\vec{s} = 25 \hat{n} = 25 \left( \dfrac{3\hat{i} - 4\hat{j}}{5} \right) = 15\hat{i} - 20\hat{j} m

The work done WW is the dot product of the net force and the displacement vector:

W=F⃗⋅s⃗=(5i^+2j^+2k^)⋅(15i^−20j^)W = \vec{F} \cdot \vec{s} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j})

W=(5×15)+(2×−20)+(2×0)W = (5 \times 15) + (2 \times -20) + (2 \times 0)

W=75−40=35W = 75 - 40 = 35 J

Answer: 3535

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