Question

8. Under a force of 10 î - 3 j + 6k newton, a body of mass 5 kg is displaced from the position 6î + 59 – 3 to the position 10î - 24 + 7k. Calculate the work done. (Ans. 121 J)​

Answer 1

Given:

• Force = 10i − 3j + 6k

• Initial position = 10i − 2j + 7k

• Final Position = 6i + 5j − 3k

Displacement is the change in position.

• Displacement = (10i − 2j + 7k) − (6i + 5j − 3k) = 4i − 7j + 10k

Formula To Be Applied:

• W = F.S

• where, W, F, and S stand for work done, Force, and displacement respectively

• In simple terms work done is the dot product of force and displacement.

Calculation:

⇒ Work Done = (10i − 3j + 6k) ∙ (4i − 7j + 10k)

⇒ Work Done = (10 × 4) + [(−3) × (−7)] + (6 × 10)

⇒ Work Done = 40 + 21 + 60

∴  Work Done = 121 Joules

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