A point of application of a force ${\rm{\vec F}} = 5\hat i - 4\hat j + 2\hat k$ is moved from $\overrightarrow {{{\rm{r}}_1}} = 2\hat i + 7\hat j + 4\hat k\,to\,\overrightarrow {{{\rm{r}}_2}} = 5\hat i + 2\hat j + 3\hat k$ the work done is

NEET UG PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1

A point of application of a force

\mathrm{\vec F} = 5\hat i - 4\hat j + 2\hat k

is moved from

\overrightarrow {{\mathrm{r}_1}} = 2\hat i + 7\hat j + 4\hat k\,to\,\overrightarrow {{\mathrm{r}_2}} = 5\hat i + 2\hat j + 3\hat k

the work done is

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Correct option: A

22 units

Option A (22 units) is correct because the work done is the dot product of the force and the displacement vector.

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{A}

2 {~m}

2\mathrm{\;m}{\mathrm{s}^{ - 1}}

x = \mathrm{m}{\mathrm{s}^{ - 1}}

\mathrm{m}{\mathrm{s}^{ - 1}}

\mathrm{\vec F} = - K\left( {y\hat i + x\hat j} \right)

\mathrm{\vec F}

9.8\;\mathrm{m}{\mathrm{s}^{ - 1}}