Question

A body has displacement (2, 4, -6) to (6, -4, 4) under a constant force $\rm{2 \hat{i}+3\hat{j}-\hat{k}}$. Find the work done.

Answer 1

Given :-

A body has displacement (2, 4, -6) to (6, -4, 4) under a constant force $\sf{2\hat{i}+3\hat{j}-\hat{k}}$.

To Find :-

The work done by the body.

Solution :-

According to the question,

$\bullet\ \sf{\overrightarrow{r_1}=2\hat{i}+4\hat{j}-6\hat{k}}$

$\bullet\ \sf{\overrightarrow{r_2}=6\hat{i}-4\hat{j}+4\hat{k}}$

$\sf{\to \Delta \overrightarrow{r}=\overrightarrow{D}=\overrightarrow {r_2}-\overrightarrow {r_1}}$

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Solving further :-

$\longrightarrow \sf{\Delta \overrightarrow{r}= 6\hat{i}-4\hat{j}+4\hat{k}-(2\hat{i}+4\hat{j}-6\hat{k}) }$

$\longrightarrow \sf{\Delta \overrightarrow{r}= 6\hat{i}-4\hat{j}+4\hat{k}-2\hat{i}-4\hat{j}+6\hat{k}}$

$\longrightarrow \sf{\Delta \overrightarrow{r}= 4\hat{i}-8\hat{j}+10\hat{k}}$

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We are given with :

$\dag\ \sf{\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}}$

We know that,

$\sf{Work\ done,\ W=\overrightarrow{F}.\overrightarrow{D}}$

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Solving further :-

$\sf{\longrightarrow W= (2\hat{i}+3\hat{j}-\hat{k}).(4\hat{i}-8\hat{j}+10\hat{k})}$

$\sf{\longrightarrow W= 8-24-10}$

$\boxed{\bf{\longrightarrow W= -26\ J}}$

OR

$\boxed{\bf{\longrightarrow W= -26\ N.m}}$

◘ Therefore, the work done by the body is -26 Joules or -26 Newton·meter.

Answer 2

$\boxed{\fcolorbox{red}{green}{verified \: \: answer}}$

ANSWER:-

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Given :

A body has displacement (2, 4, -6) to (6, -4, 4) under a constant force

$\sf{2\hat{i}+3\hat{j}-\hat{k}}2i^+3j^−k</p><p>$

To find :

Work done

Knowledge required :

Dot product of two different orthogonal vectors is equals to 0

$\sf{\hat{i}\;.\;\hat{j}=\hat{j}\;.\;\hat{k}=\hat{k}\;.\;\hat{i}=0}$

and Dot product of an orthogonal vector with itself is equals to 1

$; \sf{\hat{i}\;.\;\hat{i}=\hat{j}\;.\;\hat{j}=\hat{k}\;.\;\hat{k}=1}</p><p>$

Formulae to calculate work done

$\star\;\;\boxed{\sf{W=\vec{F}\;.\;\vec{d}}}$

[ Work done is given by the dot product of force and displacement vector ; where F is Force and d is displacement ]

Solution :

Given, position vectors of the body

$\sf{\vec{r_1}=2\;\hat{i}+4\;\hat{j}-6\;\hat{k}}r1=2i^+4j^−6k^ and$

$\sf{\vec{r_2}=6\;\hat{i}-4\;\hat{j}+4\;\hat{k}}r2=6i^−4j^+4k^o</p><p></p><p>$

Calculating displacement of body

Displacement  is known as the change in position vector.

so,

$\implies\sf{\vec{d}=\vec{r_2}-\vec{r_1}}⟹d=r2−r1$

$</p><p>\implies\sf{\vec{d}=(6\;\hat{i}-4\;\hat{j}+4\;\hat{k})-(2\;\hat{i}+4\;\hat{j}-6\;\hat{k})}⟹d=(6i^−4j^+4k^)−(2i^+4j^−6k^)$

$</p><p>\implies\sf{\vec{d}=6\;\hat{i}-4\;\hat{j}+4\;\hat{k}-2\;\hat{i}-4\;\hat{j}+6\;\hat{k}}⟹d=6i^−4j^+4k^−2i^−4j^+6k^o$

$\underline{\implies\sf{\vec{d}=4\;\hat{i}-8\;\hat{j}+10\;\hat{k}}}⟹d=4i^−8j^+10k^$

Calculating Work done by the body

Given,

$</p><p>\sf{\vec{F}=2\;\hat{i}+3\;\hat{j}-\;\hat{k}}F=2i^+3j^−k^o$

Using formula

$\implies\sf{W=\vec{F}\;.\;\vec{d}}⟹W=F.d$

$\implies\sf{W=(2\;\hat{i}+3\;\hat{j}-\;\hat{k})\;.\;(4\;\hat{i}-8\;\hat{j}+10\;\hat{k})}</p><p>$

$⟹W=(2i^+3j^−k^).(4i^−8j^+10k^)</p><p></p><p>$

Hope it's help u......