Question

A force of 10 N acting on at angle 60 degree with the horizontal direction displaces body 2 m along the surface of floor. calculate the work done?

Answer 1

Force=10N
direction=cos60d
work done=10*cos60d
=10*1/2
=5N

so work is 5N*2
=10joule

Answer 2

The angle is measured in degrees and you have provided it as 600. Which obviously means 60 since the superscript to signify degree was written as ‘0’ in your question.

Force, F = 10 N
Angle with horizontal, θ = 60°
Displacement in horizontal direction, S = 2 m
Component of force along horizontal direction is = F cosθ =(10)(cos60) =5 N 

So, work done is, W = (5)(2) = 10 J

Second case: 
Force, F = ? 
Angle with horizontal, θ = 30°
Displacement in horizontal direction, S = 2 m
Component of force along horizontal direction is=Fcosθ=(F)(cos30)=F(√3/2)                                                                                                                N

Any non­zero value of force will cause a displacement of 2 m.