Respuesta :
Answer:
The work done by worker's force is same as thermal energy 327.2 J
Explanation:
Given :
Mass of block [tex]m = 30[/tex] Kg
Displacement [tex]d = 6.50[/tex] m
Angle [tex]\theta =[/tex] 30°
Coefficient of kinetic friction [tex]\mu _{} = 0.190[/tex]
(A)
The work done by the worker's
[tex]W = Fd\cos \theta[/tex]
Here force is given by,
[tex]F \cos \theta = \mu F_{n}[/tex]
Where [tex]F_{n} = mg + F \sin \theta[/tex]
[tex]F \cos \theta = \mu (mg + F\sin \theta )[/tex]
[tex]F \cos 30 + 0.19( F\sin 30 )= \mu m g[/tex]
[tex]0.866 F + 0.095 F = 0.19 \times 30 \times 9.8[/tex]
[tex]F = \frac{55.86}{0.961}[/tex]
[tex]F = 58.13[/tex] N
So work done by,
[tex]W = 58.13 \times 6.50 \times \cos 30[/tex]
[tex]W = 327.2[/tex] J
(B)
The thermal energy of the block floor system is,
[tex]E_{th} = f_{k} d[/tex]
[tex]E_{th} = 58.13 \times 6.50 \times \cos 30[/tex]
[tex]E_{th} = 327.2 J[/tex]
Therefore, the work done by worker's force is same as thermal energy 327.2 J
