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a concrete cube of side 0.50 m and uniform density 2.0 x 103 kg m–3 is lifted 3.0 m vertically by a crane. what is the change in potential energy of the cube

Verified Answer :

Answer:

Change in potential energy = 7350 Joules

Explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube, \(d=2\times 10^3\ kg/m^3\)

The cube is lifted vertically by a crane to a height of 3 m

We know that, density \(d=\dfrac{m}{V}\)

So, m = d × V  (V = volume of cube = a³)

\(m=2\times 10^3\ kg/m^3\times (0.5\ m)^3\)

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 :

\(PE=mgh\)

PE = 250 kg × 9.8 m/s² ×3 m

PE = 7350 Joules

So, change in potential energy of the cube is 7350 Joules.

Answer:

\(U = 7357.5 J\)

Explanation:

Density of the cube is given as

\(\rho = 2.0 \times 10^3 kg/m^3\)

volume of the cube is given as

\(V = a^3\)

here we have

a = 0.50 m

so we will have

\(V = (0.50)^3\)

\(V = 0.125 m^3\)

so we will have mass of the block given as

\(mass = density \times Volume[tex]

[tex]M = 0.125 \times 2 \times 10^3\)

\(M = 0.25 \times 10^3\)

now potential energy is given as

\(U = mgh\)

\(U = 0.25 \times 10^3 \times 9.81\times 3\)

\(U = 7357.5 J\)

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