A block of mass m = 2.7 kg is pushed a distance d =4.5 m along a frictionless, h
ID: 2116271 • Letter: A
Question
A block of mass m = 2.7 kg is pushed a distance d =4.5 m along a frictionless, horizontal table by a constant applied force of magnitude F =16 N directed at an angle %u03B8 =29 below the horizontal as shown in the figure below.
(a) Determine the work done on the block by the applied force.
(b) Determine the work done on the block by the normal force exerted by the table.
(c) Determine the work done on the block by the gravitational force
(d) Determine the work done by the net force on the block.
Explanation / Answer
Here
The Work Done by Normal Component of the Force = FSin(theta)*d*Cos(90 degree)
= 16*Sin(29 degree)*4.5*Cos(90 degree)
= 0 J
Work Done by Horizontal Component = FCos(theta)*d*Cos(0 degree)
= 16*Cos(29 degree)*4.5*Cos(0 degree)
= 62.9726 J
Therefore
Work Done by Applied = 0 + 62.9726
= 62.9726 J
=============================================
Here the Normal Component = (FSin(theta) + mg )
= 2.7*9.8 + 16*Sin(29 degree)
= 34.217 N
Therefore Work Done = Fd* Cos(90 degree)
= 34.217*4.5 * Cos(90 degree)
= 0 J
========================================
Work Done by the Gravitational Force = mgd* Cos(90 degree)
= 2.7*9.8*4.5* Cos(90 degree)
= 0 J
========================================
Work Done by Net Force = = (FSin(theta)+mg)*d*Cos(90 degree)
= (16*Sin(29 degree)+2.7*9.8)*4.5*Cos(90 degree)
= 0 J
Work Done by Horizontal Component = FCos(theta)*d*Cos(0 degree)
= 16*Cos(29 degree)*4.5*Cos(0 degree)
= 62.9726 J
Therefore
Work Done by Net Force = 0 + 62.9726
= 62.9726 J
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