Question

the work done by a force 2i^ _j^+5k^ when it displaces the body from a point 3,4,6 to a point 7,2,5 is

Answer 1

Answer: 5 J


We are going to solve with concepts of vectors.

Work is defined as the dot product of Force and Displacement.

That is:

$W = \vec{F} . \vec{d}$

Here, we have:
$\vec{F} = 2\hat{\imath}-\hat{\jmath}+5\hat{k}$

Also, the body is displaced from (3,4,6) to (7,2,5). So, displacement vector would be:

$\vec{d} = (7,2,5) - (3,4,6) \\ \\ \implies \vec{d} = (7-3,2-4,5-6) \\ \\ \implies \vec{d} = (4,-2,-1) \\ \\ \implies \vec{d} = 4\hat{\imath}-2\hat{\jmath}-\hat{k}$

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For two vectors:
$\vec{a} = x_1 \hat{\imath} + y_1 \hat{\jmath} + z_1\hat{k} \\ \\ \vec{b} = x_2 \hat{\imath} + y_2 \hat{\jmath} + z_2\hat{k}$

The dot product is given as:

$\vec{a} \, .\, \vec{b} = (x_1 \hat{\imath} + y_1 \hat{\jmath} + z_1\hat{k}).(x_2 \hat{\imath} + y_2 \hat{\jmath} + z_2\hat{k}) \\ \\ \implies \vec{a} \, . \, \vec{b} = x_1x_2+y_1y_2+z_1z_2$

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Here, our data is:

$\vec{F} = 2\hat{\imath} - \hat{\jmath} + 5\hat{k} \\ \\ \vec{d} = 4\hat{\imath} - 2\hat{\jmath} -\hat{k}$

And so, work done would be:

$W = \vec{F} \, . \, \vec{d} \\ \\ \implies W = (2\hat{\imath} - \hat{\jmath} + 5\hat{k}) . (4\hat{\imath} - 2\hat{\jmath} -\hat{k}) \\ \\ \implies W = (2\times 4) + ((-1) \times (-2)) + (5 \times (-1)) \\ \\ \implies W = 8 + 2 - 5 \\ \\ \implies \boxed{\bold{W = 5 \, \, J}}$


Thus, Work done is 5 joules.