Question

worksheet 78

1. a stone is dropped into a lake. calculate the increase in pressure on the stone caused by the water when it sinks from 2 m deep to 12 m deep. (the density of water is 1,000 kg/m³ and gravitational field strength is 9.8 n/kg).
2. the density of water is 1000 kg/m³. calculate the pressure at the bottom of a dam 6 m deep. (gravitational field strength = 9.8 n/kg).
3. an elephant has a mass of 2000 kg. the surface area of one of its foot is 1000cm² how much pressure does it exert on the ground when standing?

Explanation:

Step1: Recall pressure - depth formula

The formula for pressure due to a fluid column is $P =
ho gh$, where $
ho$ is the density of the fluid, $g$ is the gravitational field strength and $h$ is the depth.

Step2: Solve question 1

For the stone in the lake:
The initial depth $h_1 = 2m$ and final depth $h_2=12m$.
The pressure at depth $h_1$ is $P_1=
ho gh_1$ and at depth $h_2$ is $P_2=
ho gh_2$.
The increase in pressure $\Delta P=P_2 - P_1=
ho g(h_2 - h_1)$.
Substitute $
ho = 1000kg/m^3$, $g = 9.8N/kg$ and $h_2 - h_1=(12 - 2)m = 10m$.
$\Delta P=1000\times9.8\times10=98000Pa$.

Step3: Solve question 2

For the dam:
Using the formula $P=
ho gh$, with $
ho = 1000kg/m^3$, $g = 9.8N/kg$ and $h = 6m$.
$P=1000\times9.8\times6 = 58800Pa$.

Step4: Solve question 3

First, convert the surface - area of one foot from $cm^2$ to $m^2$. Since $1m = 100cm$, $1cm^2=10^{- 4}m^2$, so $A = 1000cm^2=0.1m^2$.
The weight of the elephant $F = mg$, where $m = 2000kg$ and $g = 9.8N/kg$. So $F=2000\times9.8 = 19600N$.
The elephant has 4 feet. The total force on the ground is $F$. The pressure $P=\frac{F}{4A}$.
Substitute $F = 19600N$ and $A = 0.1m^2$.
$P=\frac{19600}{4\times0.1}=49000Pa$.

Answer:

1. $98000Pa$
2. $58800Pa$
3. $49000Pa$