Question

2. A body moves from a position r2- (2î-3ſ-4k) m to a position, r2-(31-439-5k) m under the influence of a constant force F=(41 - 4ſ + 5k) N. The work done by the force is (A) 57J (B) 58J (C) 59J (D) 60J​

Answer 1

✦Appropriate Question:-

A body moves from a position r₁ = (2i-cap-3j-cap-4k-cap) m to a position, r₂-(3i-cap-4j-cap-5k-cap) m under the influence of a constant force F=(4i-cap - j-hat + 6k) N. The work done by the force is

(A) 57J

(B) 58J

(C) 59J

(D) 60J​

✦Solution:-

Work done is defined as the product of magnitude force to the displacement moved by the particle due to the application of the force.

We are supposed to find the displacement,

Displacement = Final Position - Initial Position

$\\\quad\mapsto\quad\sf D = \bigg(3\hat{i}-4\hat{j}+5\hat{k}\bigg)-\bigg(2\hat{i}-3\hat{j}-4\hat{k}\bigg)$

$\\\quad\mapsto\quad\sf D = \bigg[(3-2)\hat{i}+(-4+3)\hat{j}+(5+4})\hat{k}\bigg]$

$\\\quad\mapsto\quad\boxed{\sf D = \hat{i}-\hat{j}+9\hat{k}}\bigstar$

Now, apply the following formula and calculate the work done.

$\\\quad\dashrightarrow\quad \sf Work\:done=\overrightarrow{F}.\overrightarrow{S}$

$\\\quad\dashrightarrow\quad \sf Work\:Done= \bigg(4\hat{i}+\hat{j}+6\hat{k}\bigg).\bigg(\hat{i}-\hat{j}+9\hat{k}\bigg)$

$\\\quad\dashrightarrow\quad\sf Work\:Done=(4)(1)+(1)(-1)+(6)(9)$

$\\\quad\dashrightarrow\quad\sf Work\:Done=4-1+54$

$\quad\therefore\quad\boxed{\boxed{\sf Work\:Done=57 J\:(a)}}\bigstar$

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