Question

A body when acted by a force of 10kgf is displaced by 0.5m. Calculate the work done by the force in each case,when displacement is:in the direction of drive ,at an angle of 60° and normal to the force.

Answer 1

Answer:

W=F.S.cos∅

Explanation:

a) When displacement is in direction of drive

W=F.S.cos∅=10x0.5xcos0 = 5N

b) At an angle 60,

W=F.S.cos∅=10x0.5xcos60 =10x0.5x0.5 = 2.5N

c) Normal to the force

W=F.S.cos∅=10x0.5xcos90 = 0 N

Hope it helps :)