Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, [tex]d=2\times 10^3\ kg/m^3[/tex]
The cube is lifted vertically by a crane to a height of 3 m
We know that, density [tex]d=\dfrac{m}{V}[/tex]
So, m = d × V (V = volume of cube = a³)
[tex]m=2\times 10^3\ kg/m^3\times (0.5\ m)^3[/tex]
m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :
[tex]PE=mgh[/tex]
PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
Answer:
[tex]U = 7357.5 J[/tex]
Explanation:
Density of the cube is given as
[tex]\rho = 2.0 \times 10^3 kg/m^3[/tex]
volume of the cube is given as
[tex]V = a^3[/tex]
here we have
a = 0.50 m
so we will have
[tex]V = (0.50)^3[/tex]
[tex]V = 0.125 m^3[/tex]
so we will have mass of the block given as
[tex]mass = density \times Volume[tex]
[tex]M = 0.125 \times 2 \times 10^3[/tex]
[tex]M = 0.25 \times 10^3[/tex]
now potential energy is given as
[tex]U = mgh[/tex]
[tex]U = 0.25 \times 10^3 \times 9.81\times 3[/tex]
[tex]U = 7357.5 J[/tex]
