Question

Calculate in joule the magnitude of the work needed to be done to lift 50 cubic meters of water to a height of 10 meters?

Answer 1

I got `98,000` joules.

Explanation:

We got `50 \ "m"^3` of water. We first convert it to `"cm"^3`.

We know that `1 \ "m"^3=1,000,000 \ "cm"^3=10^6 \ "cm"^3`.

Also, the density of water is `1 \ "g/cm"^3`. So the mass of water here is:

`10^6color(red)cancelcolor(black)("cm"^3)*(1 \ "g")/(color(red)cancelcolor(black)("cm"^3))=10^6 \ "g"`

Now, we convert it to kilograms so it's easier to work with. Know that, `1 \ "kg"=10^3 \ "g"`. And so,

`10^6color(red)cancelcolor(black)"g"*(1 \ "kg")/(10^3color(red)cancelcolor(black)"g")=10^3 \ "kg"`

Work done is given by the equation:

`W=F*d`

• `F` is the force in newtons

• `d` is the distance in meters

Here, the force is the weight of the water, which is: `10^3 \ "kg"*9.8 \ "m/s"^2=9.8*10^3 \ "N"`.

Therefore, the work done is:

`W=9.8*10^3 \ "N"*10 \ "m"`

`=9.8*10^4 \ "J"`

`=98,000 \ "J"`

Now that's a lot of work!