Respuesta :
Answer
given,
Fā is horizontal = 40 N
Fā is normal = 20 N
Fā is parallel = 30 N
work done by
Wā = 0 as force is acting perpendicular to the direction of motion.
as the motion moved to 0.8 cm
Wā = Fā x d
Wā = 30 x 0.8
Wā = 24 J
Wā = Fā x d
Wā = Fā cos ā x d
Wā = 40 cos 30ā° x 0.8
Wā = 27.21 J
Answer:
The work done by the forces on the block [tex]W_{1} =27.21J[/tex]
[tex]$\mathrm{WF}_{2} =8J$[/tex]
[tex]$\mathrm{WF}_3} =24J$[/tex]
Explanation:
Given,
[tex]$F_{1}$[/tex]is horizontal [tex]$=40 \mathrm{~N}$[/tex]
[tex]$\mathrm{F}_{2}$[/tex] is normal [tex]$=20 \mathrm{~N}$[/tex]
[tex]$\mathrm{F}_{3}$[/tex] is parallel [tex]$=30 \mathrm{~N}$[/tex]
The motion moved to [tex]$0.8 \mathrm{~cm}$[/tex]
[tex]$W_{1}=F_{1} \times d$[/tex]
[tex]$W_{1}=F_{1} \cos \emptyset \times d$[/tex]
[tex]$W_{1}=40 \cos 30^{\circ} \times 0.8$[/tex]
[tex]$W_{1}=27.21 \mathrm{~J}$[/tex]
[tex]$\mathrm{WF}_{2} =20(\sin 30)(0.8)=8J$[/tex]
[tex]$\mathrm{WF}_{3} =30(0.8)=24J$[/tex]
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