Constant forces = 2î - 5ĵ + 6k̂ and = -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 and respectively, then the vector .

For two vectors and at an angle θ to each other:

Dot Product is defined as: .
Resultant Vector is equal .
Work: The work (W) done by a force () in moving (displacing) an object along a vector is given by: W = .

Calculation:

Let's say that the forces acting on the particle are = 2î - 5ĵ + 6k̂ and = -î + 2ĵ - k̂.

∴ The resulting force acting on the particle will be .

⇒ = (2î - 5ĵ + 6k̂) + (-î + 2ĵ - k̂)

⇒ = î - 3ĵ + 5k̂.

Since the particle is moved from the point 4î - 3ĵ - 2k̂ to the point 6î + ĵ - 3k̂, the displacement vector will be:

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

⇒ = 2î + 4ĵ - k̂.

And finally, the work done W will be:

W = = (î - 3ĵ + 5k̂).(2î + 4ĵ - k̂)

⇒ W = (1)(2) + (-3)(4) + (5)(-1)

⇒ W = 2 - 12 - 5 =

∴ -15 units.