Question

1. A college student earning some summer money pushes a lawn mower on a level lawn with a constant force of 450 N at an angle of 20° downward from the horizontal. How far does the student push the mower in doing 2.5x10³ J of work?  

Answer 1

Using Above FBD:

Net force in horizontal direction will be:

Fx_net = F_app*cos

F_app = Force applied by student = 450 N

= Angle = 20 deg

Now work-done is given by:

W = F*d

Where F is force applied in the direction of displacement, So

W = F_app*d*cos

d = W/(F_app*cos )

Using given values:

d = 2.5*10^3/(450*cos 20 deg)

d = displacement of mower = 5.9121 m

In two significant figures

d = 5.9 m

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