The graph shows the horizontal component of the net force ( F net cos ? , where

Question

The graph shows the horizontal component of the net force (Fnetcos? , where ?  is the angle between the net force vector and the horizontal) that acts on a 5.9kg  block as it moves along a flat horizontal surface.

a) Find the net work done on the block as it moves from s=0  to s=10m .

b) Find the final speed of the block if it starts from rest at s = 0

s=0

The graph shows the horizontal component of the net force (Fnetcos? , where ? is the angle between the net force vector and the horizontal) that acts on a 5.9kg block as it moves along a flat horizontal surface. Find the net work done on the block as it moves from s=0 to s=10m . Find the final speed of the block if it starts from rest at s = 0 s=0

Explanation / Answer

a.)Work done=Fnet.A=Fnetcos(theta)*A=Area of the graph=16-4=12 J

b.)Work done=gain in KE=0.5mv^2

12=0.5*5.9*v^2

So v=2.017 m/s

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