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 3.7kg  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 .

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 3.7kg 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 .

Explanation / Answer

a) Area under the curve of F-S gives Work done. here there are two rectangles so areas are

W1=4*3=12 J and W2=(-3)*2=-6J hence W=12-6= 6J (ans)

b)Workdone=change in kinetic energy ==> W=1/2mv^2-0

6=(1/2*3.7*v^2) ==>v=1.8 m/s

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