Question

a body of mass 5 kg moves from position 6i + 5j - 3k metre to a position 10i - 2j +7k metre due to a force 10i - 3j + 6k N. calculate the work done.

Answer 1

F = 10i-3j+6k
displacement = (10i-2j+7k) - (6i+5j-3k)
= (4i-7j+10k)
work done =force.displacement
= (10i-3j+6k).(4i-7j+10k)
= 10×4+3×7+6×10
= 40 + 21 +60 = 121 Joules

Answer 2

Work equals the dot product of force and displacement.

W=F∙d,

Force is given as 10i−3j+6k, F = 10i-3j+6k,

displacement = (10i-2j+7k) - (6i+5j-3k), = (4i-7j+10k),

work done =force.

Displacement = (10i-3j+6k).(4i-7j+10k) = 10×4+3×7+6×10 = 40 + 21 +60 = 121 Joules.

W=(10i−3j+6k)∙(4i−7j+10k),

W=(10∗4)+((−3)∗(−7))+(6∗10)=40+21+60=121.

Assuming positions given were in meters, work is 121 Joules.