# Science:Math Exam Resources/Courses/MATH101/April 2015/Question 09 (b)/Solution 1

First, calculate the work required to move the iron plate at the height ${\displaystyle zm}$ with the width ${\displaystyle dz}$.

Since we move from the 2m basement below ground level, the distance that the object move is ${\displaystyle \color {OrangeRed}d=(z+2)m}$.

On the other hand, since the plate at the height ${\displaystyle z}$ is a square with the side length ${\displaystyle (3-z)m}$, its volume is ${\displaystyle V=(3-z)^{2}dzm^{3}}$.

Using the formula for the mass, this gives the mass of the plate: ${\displaystyle m=\rho \cdot V=(3-z)^{2}dzm^{3}\cdot 8000kg/m^{3}=8000\cdot (3-z)^{2}dzkg}$.

Then, from the formula for the force, we have ${\displaystyle {\color {OliveGreen}{F}}=mg=8000\cdot (3-z)^{2}dzkg\cdot 9.8m/s^{2}={\color {OliveGreen}9.8\cdot 8000\cdot (3-z)^{2}dz(kg\cdot m/s^{2})}}$.

Therefore, the required work to move the iron plate at the height ${\displaystyle zm}$ is ${\displaystyle W={\color {OliveGreen}F}{\color {OrangeRed}d}=9.8\cdot 8000\cdot (3-z)^{2}dz(kg\cdot m/s^{2})\cdot (z+2)m=9.8\cdot 8000\cdot (3-z)^{2}(z+2)dz(kg\cdot m^{2}/s^{2})}$

Finally, since the range of ${\displaystyle z}$ is from ${\displaystyle 0}$ to ${\displaystyle 3}$, we integrate in this range and get the total required work to move all of the iron from its starting position to its present position:

${\displaystyle W=\color {blue}\int _{0}^{3}9.8\cdot 8000\cdot (3-z)^{2}(z+2)dz(kg\cdot m^{2}/s^{2})}$.