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2 years ago

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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

2 answers:

Alex787 [66]2 years ago

8 0

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

$U = mgh$

$U = 0.25 \times 10^3 \times 9.81\times 3$

$U = 7357.5 J$

AlladinOne [14]2 years ago

5 0

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.

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Hi there!

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The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

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v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

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