Class 9 - Unit # 5 : Forces And Matter - Solved Numericals

Hooke's Law

4. Calculate the spring constant for a spring which extends by a distance of $3.5\text{ cm}$ when a load of $14\text{ N}$ is hung from its end.

Data:

• Load Hung ($F$) = $14\text{ N}$
• Extension produced ($x$) = $3.5\text{ cm} = \frac{3.5}{100}\text{ m} = 0.035\text{ m}$
• Spring Constant ($k$) = ?

Solution:

According to Hooke’s Law:

$$F = k \cdot x$$

Rearranging the formula to solve for the spring constant ($k$):

$$k = \frac{F}{x}$$ $$k = \frac{14\text{ N}}{0.035\text{ m}}$$ $$k = 400\text{ N/m}$$

RESULT: The spring constant for the spring is $400\text{ N/m}$.

Pressure

8. A boy is pressing a thumbtack into a piece of wood with a force of $20\text{ N}$. The surface area of the head of the thumbtack is $1\text{ cm}^2$ and the cross-sectional area of the tip of the thumbtack is $0.01\text{ cm}^2$. Calculate:
a) The pressure exerted by the boy's thumb on the head of the thumbtack.
b) The pressure of the tip of the thumbtack on the wood.
c) What conclusion can be drawn from the answers of part (a) and (b)?

Data:

• Force applied ($F$) = $20\text{ N}$
• Area of Head ($A_1$) = $1\text{ cm}^2 = 1 \times 10^{-4}\text{ m}^2 = 0.0001\text{ m}^2$
• Area of Tip ($A_2$) = $0.01\text{ cm}^2 = 0.01 \times 10^{-4}\text{ m}^2 = 1 \times 10^{-6}\text{ m}^2$
• a) Pressure applied on head ($P_1$) = ?
• b) Pressure applied by tip ($P_2$) = ?

Solution:

According to the definition of pressure ($P = \frac{F}{A}$):

a) Pressure applied on the head ($P_1$):

$$P_1 = \frac{F}{A_1}$$ $$P_1 = \frac{20\text{ N}}{1 \times 10^{-4}\text{ m}^2}$$ $$P_1 = 200,000\text{ Pa} \quad (\text{or } 2 \times 10^5\text{ N/m}^2)$$

b) Pressure applied by the tip on the wood ($P_2$):

$$P_2 = \frac{F}{A_2}$$ $$P_2 = \frac{20\text{ N}}{1 \times 10^{-6}\text{ m}^2}$$ $$P_2 = 20,000,000\text{ Pa} \quad (\text{or } 2 \times 10^7\text{ N/m}^2)$$

c) Conclusion:

$$\frac{P_2}{P_1} = \frac{20,000,000}{200,000} = 100$$

RESULT: a) The pressure applied on the head is $200,000\text{ Pa}$. b) The pressure applied by the tip on the wood is $20,000,000\text{ Pa}$. c) We conclude that for the same force, a smaller area experiences a much higher pressure; the tip exerts 100 times more pressure on the wood than the thumb exerts on the head.

9. A basic hydraulic system has small and large pistons of cross-sectional areas $0.005\text{ m}^2$ and $0.1\text{ m}^2$ respectively. A force of $20\text{ N}$ is applied to the small piston. Calculate:
a) The pressure transmitted into the hydraulic fluid.
b) The force acting at the large piston.
c) Discuss the distance travelled by the small and large pistons.

Data:

• Force applied on small piston ($F_1$) = $20\text{ N}$
• Area of small piston ($A_1$) = $0.005\text{ m}^2$
• Area of large piston ($A_2$) = $0.1\text{ m}^2$
• a) Pressure transmitted to fluid ($P$) = ?
• b) Force at the large piston ($F_2$) = ?

Solution:

a) According to Pascal's Principle, pressure applied to a fluid is transmitted equally throughout it:

$$P = \frac{F_1}{A_1}$$ $$P = \frac{20\text{ N}}{0.005\text{ m}^2} = 4000\text{ Pa}$$

b) The force acting at the large piston ($F_2$):

$$P = \frac{F_2}{A_2} \implies F_2 = P \times A_2$$ $$F_2 = 4000\text{ Pa} \times 0.1\text{ m}^2$$ $$F_2 = 400\text{ N}$$

c) Distance analysis (Work Conservation):

Since the volume of fluid displaced must remain constant ($\Delta V = A_1 \cdot d_1 = A_2 \cdot d_2$), the ratio of distances traveled is inversely proportional to the cross-sectional areas:

$$\frac{d_1}{d_2} = \frac{A_2}{A_1} = \frac{0.1}{0.005} = 20$$

RESULT: a) The pressure transmitted to the fluid is $4000\text{ Pa}$. b) The force generated at the large piston is $400\text{ N}$. c) To maintain work conservation, the small piston moves a much longer distance ($20\text{ times}$ more) while the large piston moves a proportionally shorter distance.