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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

User JasonE
by
8.5k points

2 Answers

2 votes

Answer:


U = 7357.5 J

Step-by-step explanation:

Density of the cube is given as


\rho = 2.0 * 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 * Volume[tex]</p><p>[tex]M = 0.125 * 2 * 10^3


M = 0.25 * 10^3

now potential energy is given as


U = mgh


U = 0.25 * 10^3 * 9.81* 3


U = 7357.5 J

User Sbking
by
7.7k points
3 votes

Answer:

Change in potential energy = 7350 Joules

Step-by-step explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube,
d=2* 10^3\ kg/m^3

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

We know that, density
d=(m)/(V)

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


m=2* 10^3\ kg/m^3* (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.

User Anslem
by
8.9k points
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