Constant forces \(\rm \vec P\) = 2î - 5ĵ + 6k̂ and \(\rm \vec Q\) = -î + 2ĵ - k̂ act on a particle. The work done when the particle is displaced from A whose position vector is 4î - 3ĵ - 2k̂, to B whose position vector is 6î + ĵ - 3k̂, is:

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Question

This question was previously asked in

NIMCET 2014 Official Paper

Answer 1

Concept:

If two points A and B have position vectors \(\rm \vec A\) and \(\rm \vec B\) respectively, then the vector \(\rm \vec {AB}=\vec B-\vec A\).

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

Dot Product is defined as: \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
Resultant Vector is equal \(\rm \vec A + \vec B\).
Work: The work (W) done by a force (\(\rm \vec F\)) in moving (displacing) an object along a vector \(\rm \vec D\) is given by: W = \(\rm \vec F.\vec D=|\vec F||\vec D|\cos \theta\).

Calculation:

Let's say that the forces acting on the particle are \(\rm \vec P\) = 2î - 5ĵ + 6k̂ and \(\rm \vec Q\) = -î + 2ĵ - k̂.

∴ The resulting force acting on the particle will be \(\rm \vec F=\vec P+\vec Q\).

⇒ \(\rm \vec F\) = (2î - 5ĵ + 6k̂) + (-î + 2ĵ - k̂)

⇒ \(\rm \vec F\) = î - 3ĵ + 5k̂.

Since the particle is moved from the point 4î - 3ĵ - 2k̂ to the point 6î + ĵ - 3k̂, the displacement vector \(\rm \vec D\) will be:

\(\rm \vec D=\vec{AB}=\vec B-\vec A\)

= (6î + ĵ - 3k̂) - (4î - 3ĵ - 2k̂)

⇒ \(\rm \vec D\) = 2î + 4ĵ - k̂.

And finally, the work done W will be:

W = \(\rm \vec F.\vec D\) = (î - 3ĵ + 5k̂).(2î + 4ĵ - k̂)

⇒ W = (1)(2) + (-3)(4) + (5)(-1)

⇒ W = 2 - 12 - 5 =

∴ -15 units.