Pressure Worksheet | PDF - Free Printable

Let's solve each problem step by step.

---

Problem 1:

Find the pressure exerted on the floor by a 100 N box whose bottom area is 40 cm by 50 cm. (Use Pa for units)

#### Solution:
The formula for pressure is:
\[
P = \frac{F}{A}
\]
where:
- \( P \) is the pressure,
- \( F \) is the force (weight of the box),
- \( A \) is the area over which the force is applied.

1. Given:
- Force (\( F \)) = 100 N
- Dimensions of the bottom area = 40 cm × 50 cm

2. Calculate the area (\( A \)):
\[
A = 40 \, \text{cm} \times 50 \, \text{cm} = 2000 \, \text{cm}^2
\]
Convert the area to square meters (\( \text{m}^2 \)):
\[
1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \quad \Rightarrow \quad A = 2000 \, \text{cm}^2 \times 10^{-4} = 0.2 \, \text{m}^2
\]

3. Calculate the pressure:
\[
P = \frac{F}{A} = \frac{100 \, \text{N}}{0.2 \, \text{m}^2} = 500 \, \text{Pa}
\]

#### Answer:
\[
\boxed{500 \, \text{Pa}}
\]

---

Problem 2:

Find the pressure exerted by a waterbed with dimensions of 2 m × 2 m which is 30 cm thick. (Hint: Use density of water)

#### Solution:
The pressure at the bottom of the waterbed is due to the weight of the water. The formula for pressure due to a fluid column is:
\[
P = \rho g h
\]
where:
- \( P \) is the pressure,
- \( \rho \) is the density of the fluid (water),
- \( g \) is the acceleration due to gravity,
- \( h \) is the height (or thickness) of the water column.

1. Given:
- Density of water (\( \rho \)) = 1000 kg/m³
- Acceleration due to gravity (\( g \)) = 9.8 m/s²
- Thickness of the waterbed (\( h \)) = 30 cm = 0.3 m

2. Calculate the pressure:
\[
P = \rho g h = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 0.3 \, \text{m} = 2940 \, \text{Pa}
\]

#### Answer:
\[
\boxed{2940 \, \text{Pa}}
\]

---

Problem 3:

Find the pressure exerted on the ground (in psi) by a 250 lb person who is sitting on a chair whose 4 legs touch the ground with an area of 1" × 2" each.

#### Solution:
The formula for pressure is:
\[
P = \frac{F}{A}
\]
where:
- \( P \) is the pressure,
- \( F \) is the force (weight of the person),
- \( A \) is the total area over which the force is applied.

1. Given:
- Weight of the person (\( F \)) = 250 lb
- Area of each leg = 1" × 2" = 2 in²
- Number of legs = 4

2. Calculate the total area (\( A \)):
\[
A = 4 \times 2 \, \text{in}^2 = 8 \, \text{in}^2
\]

3. Calculate the pressure in pounds per square inch (psi):
\[
P = \frac{F}{A} = \frac{250 \, \text{lb}}{8 \, \text{in}^2} = 31.25 \, \text{psi}
\]

#### Answer:
\[
\boxed{31.25 \, \text{psi}}
\]

---

Problem 4:

Find the pressure exerted on a 6 ft × 10 ft wall by the atmosphere, if Atmospheric pressure is 14.7 psi.

#### Solution:
The atmospheric pressure is already given as 14.7 psi. Since pressure is uniform and does not depend on the area of the wall, the pressure exerted on the wall is simply the atmospheric pressure.

#### Answer:
\[
\boxed{14.7 \, \text{psi}}
\]

---

Problem 5:

Find the pressure exerted by the atmosphere on a car window which is 30 cm tall and 40 cm wide if atmospheric pressure is 101,000 Pa. (Remember that 1 Pa = N/m²)

#### Solution:
The atmospheric pressure is already given as 101,000 Pa. Since pressure is uniform and does not depend on the size of the window, the pressure exerted on the window is simply the atmospheric pressure.

#### Answer:
\[
\boxed{101000 \, \text{Pa}}
\]

---

Final Answers:

1. \(\boxed{500 \, \text{Pa}}\)
2. \(\boxed{2940 \, \text{Pa}}\)
3. \(\boxed{31.25 \, \text{psi}}\)
4. \(\boxed{14.7 \, \text{psi}}\)
5. \(\boxed{101000 \, \text{Pa}}\)