Question

A particale moves from position r1= 3i+2j-6k to position r2= 14i+13j+5k under the action of force f= 4i+j+3k. The work done by the force is,

Answer 1

Heya user !!!

Here's the answer you are looking for

R1 = 3i+2j-6k

R2 = 14i+13j+5k

So their displacement vector, say S = R2 - R1

= 14i+13j+5k - (3i+2j-6k)

= 14i + 13j + 5k - 3i - 2j + 6k

= 11i + 11j + 11k

Force = 4i+j+3k

♦ Work done = F . s

= (4i+j+3k) . (11i + 11j + 11k)

= 4.11 + 1.11 + 3.11

= 44 + 11 + 33

=88 J

Therefore, the work done is 88J


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Answer 2

(r₂ - r₁) = (14i + 13j + 5k) - (3i + 2j - 6k)
= 11i + 11j + 11k

W = F · (r₂ - r₁)
= (4i + j + 3k) · (11i + 11j + 11k)
= 44 + 11 + 33
= 88 joules

Work done is 88 joules