Question

1. A uniform force of (2i+j) N acts on a particle of mass 1 kg. The particle displaces from position (3j+k) m to (5i +3j) m. The work done by the force on the particle is (a) 9 J (b) 6J c) 10 J (d) 12 J​

Answer 1

${\textsf{\textbf{\underline{Given :-}}}}$

• Force = 2i + j
• Mass = 1 kg
• Initial position = 3j + k
• Final position = 5i + 3j

${\textsf{\textbf{\underline{To\;Find :-}}}}$

Work done

${\textsf{\textbf{\underline{Solution :-}}}}$

First of all, we need to find the distance of the particle.

S = D₂ - D₁

• S = Distance
• D₂ = Final position
• D₁ = Initial position

⇒ S = (5i + 3j) - (3j + k)

⇒ S = 5i + 3j - 3j - k

• On cancelling 3j

⇒ S = 5i - k

Work done is defined as the product of the force applied and displacement/distance covered by the particle.

W = F × s

• W = Work done
• F = Force
• s = displacement

⇒ W = (2i + j).(5i - k)

⇒ W = (2 × 5).(1 × 1)

⇒ W = 10 J

Option C is correct.