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NVAEasyJEE 2023Work Done by Force

JEE Physics 2023 Question with Solution

A small particle moves to position 5i^−2j^+k^5\hat{i} - 2\hat{j} + \hat{k} from its initial position 2i^+3j^−4k^2\hat{i} + 3\hat{j} - 4\hat{k} under the action of force 5i^+2j^+7k^5\hat{i} + 2\hat{j} + 7\hat{k} N. The value of work done will be _____ J.

Answer

Correct answer:40

Step-by-step solution

Standard Method

Given: Initial position is 2i^+3j^−4k^2\hat{i} + 3\hat{j} - 4\hat{k}, final position is 5i^−2j^+k^5\hat{i} - 2\hat{j} + \hat{k}, and force is 5i^+2j^+7k^5\hat{i} + 2\hat{j} + 7\hat{k} N.

Find: Work done.

Use the dot product formula for work:

W=F⃗⋅(rf⃗−ri⃗)W = \vec{F} \cdot (\vec{r_f} - \vec{r_i})

From the given working,

W=(5i^+2j^+7k^)⋅((5i^−2j^+k^)−(2i^+2j^−4k^))W = (5\hat{i}+2\hat{j}+7\hat{k}) \cdot \left((5\hat{i}-2\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}-4\hat{k})\right)

Thus,

W=40 JW = 40 \, \text{J}

Therefore, the work done is 40 J40 \, \text{J}.

Vector Displacement Approach

Given: Force and initial and final position vectors are provided.

Find: The numerical value of work done.

Work done by a constant force equals force dotted with displacement.

The displacement vector is obtained from final position minus initial position. Then take the dot product with the force vector to get the scalar work.

From the extracted solution, the final computed result is 40 J40 \, \text{J}. Hence the required numerical answer is 40.

Common mistakes

  • Using the position vector directly instead of displacement is incorrect because work depends on change in position, not absolute position. Always compute rf⃗−ri⃗\vec{r_f}-\vec{r_i} first.

  • Taking magnitude multiplication instead of dot product is wrong because work is F⃗⋅s⃗\vec{F} \cdot \vec{s}, not merely ∣F⃗∣ ∣s⃗∣|\vec{F}|\,|\vec{s}|. Use component-wise multiplication and addition.

  • Missing signs of the j^\hat{j} and k^\hat{k} components changes the answer. Keep careful track of negative components while subtracting vectors.

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