A particle of mass 0.5 kg is displaced from position r1vector(2,3,1) to r2 vector(4,3,2) by applying a force magnitude 30N which is acting along (icap+jcap+kcap).The work done by the force is
Answers
W = 30/root 3* (i+j+k). (2i+k)=10root3*(2+1)=30root3 jule
Given : A particle of mass 0.5 kg is displaced from position r₁ (2,3,1) to r₂ (4,3,2) by applying a force magnitude 30N which is acting along (i + j + k).
To find : The work done by the force.
solution : force of magnitude 30N is acting along (i + j + k).
unit vector of force, n = (i + j + k)/√(1² + 1² + 1²)
= (1/√3)i + (1/√3)j + (1/√3)k
so, force vector, F = n|F|
= [(1/√3)i + (1/√3)j + (1/√3)k]30
= (10√3)i + (10√3)j + (10√3)k
position Vector, r = r₂ (4, 3, 2) - r₁ (2, 3, 1)
= (4 - 2)i + (3 - 3)j + (2 - 1)k
= 2i + k
The workdone by the force, W = F.r
= [(10√3)i + (10√3)j + (10√3)k ].[2i + k]
= (10√3).(2) + (10√3)
= 30√3 J
Therefore the workdone by the force is 30√3 J.
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