Question

A force of 10 N acting on a body at an angle of 60°with the horizontal direction displaces the body through a distance of 2 mts along the surface of a floor.calculate the work done.now let the force on the body makes an angle of 30° with the horizontal.What is the value of the force to displace the body through 2 mt along the surface of the floor.

Answer 1

First Case
Work = 10*2*.5 = 10 J

Answer 2

Hey dear,

◆ Answer-

W = 10 J

F2 = 8.66 N

◆ Explaination-

# Given-

F = 10 N

s = 2 m

θ1 = 60°

θ2 = 30°

# Solution-

Initially, horizontal component of force,

F1 = Fcosθ1

F1 = 10 × cos60°

F1 = 10 × 0.5

F1 = 5 N

Work done by this force for displacement -

W = F1 × s

W = 5 × 2

W = 10 J

Later, horizontal component of force,

F2 = Fcosθ2

F2 = 10 × cos30°

F2 = 10 × 0.866

F2 = 8.66 N

Hope this helps...