Respuesta :
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]
thus,
volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height
thus,
center of mass lies at, h' = [tex]\frac{1}{4}\times4=1m[/tex]
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.
Given to us
Height of the cone = 4m
The radius of the cone = 1.2 m
Density of the material = 600 kg/m³
We know that the work gone against the gravity is given as,
[tex]W = mgh'[/tex]
where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.
Also, the mass of an object is the product of its volume and density, therefore,
[tex]m = v \times \rho[/tex]
We know the value of the volume of a cone,
[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]
The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,
[tex]h' = \dfrac{h}{4}[/tex]
Substitute all the values we will get,
[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]
Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.
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