A block of mass m = 2.50 kg is pushed d = 3.00 m along a frictionless horizontal table by a constant applied force of magnitude F = 18.0 N directed at an angle θ = 25.0° below the horizontal as shown in the figure below.A block labeled m is on a horizontal surface. An arrow labeled vector F points downward and to the right at an angle θ above the horizontal, and acts upon the upper left corner of the block. A faded image of the block is a distance d to the right of the block. (a) Determine the work done by the applied force. (b) Determine the work done by the normal force exerted by the table. (c) Determine the work done by the force of gravity. (d) Determine the work done by the net force on the block.

Question
Asked Nov 14, 2019
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A block of mass 

m = 2.50 kg

 is pushed 

d = 3.00 m

 along a frictionless horizontal table by a constant applied force of magnitude 

F = 18.0 N

 directed at an angle 

θ = 25.0°

 below the horizontal as shown in the figure below.

A block labeled m is on a horizontal surface. An arrow labeled vector F points downward and to the right at an angle θ above the horizontal, and acts upon the upper left corner of the block. A faded image of the block is a distance d to the right of the block.
 
(a) Determine the work done by the applied force.
 

(b) Determine the work done by the normal force exerted by the table.
 

(c) Determine the work done by the force of gravity.
 

(d) Determine the work done by the net force on the block.
 
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Expert Answer

Step 1
Gfven
Mass of the block, m2.50 kg
Displacement, d = 3.00m
Applred foce Fopphied
Angle of foore
Accelesatim due to pavty, g= 9.81 m/s2
250
F
Horifantal
frictionless
Aurfae
-d
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Gfven Mass of the block, m2.50 kg Displacement, d = 3.00m Applred foce Fopphied Angle of foore Accelesatim due to pavty, g= 9.81 m/s2 250 F Horifantal frictionless Aurfae -d

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Step 2
bee body dliagram of block
FCos
Fne
(4) woxk done by appliedl foste.
Wepp = Fapplnd)(d) coso
- (l8.0 N) (3.00m) Coa C25.0)
= 48.94 J
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Image Transcriptionclose

bee body dliagram of block FCos Fne (4) woxk done by appliedl foste. Wepp = Fapplnd)(d) coso - (l8.0 N) (3.00m) Coa C25.0) = 48.94 J

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Step 3
Noo mal foarie act on th loch
fom the Auface N
N-mg - Fsin=o
Efy0
ma t FAino
N
= (2.50 19) ( 9.81 m/s)+ ({p.0N)Sin (25)
32.13 N
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Noo mal foarie act on th loch fom the Auface N N-mg - Fsin=o Efy0 ma t FAino N = (2.50 19) ( 9.81 m/s)+ ({p.0N)Sin (25) 32.13 N

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