Calculate the work done by a person carrying a load of 30kg when he walks a distance of 10m in (1)the horizontal direction (2)a direction inclined at 30° with the horizontal (3)the vertical direction.
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Explanation:
Consider the following,
work done is written as W = F.s = FsCosθ here F is the force applied, s is the displacement caused such that
F = mg = 30 X 10 = 300 N s
W( horizontal direction)= FsCos(0)=3000J , θ is the angle between (vectors) F and s now,
as F and s are perpendicular for vertical direction, θ = 90 thus, W = FsCos90 = 0
W=Fscos30=2598J
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