NLM O15

Solution 15

Solution to Question 15

Step 1: Calculate Segment Stiffness

The total spring constant is $k = 1.00 \, \text{N/cm} = 100 \, \text{N/m}$.
The total length is $L = 0.4 \, \text{m} + 0.1 \, \text{m} = 0.5 \, \text{m}$.
Point P divides the spring into two segments:

Left Segment ($L_1 = 0.4$ m): $k_1 = \frac{k L}{L_1} = 100 \times \frac{0.5}{0.4} = 125 \, \text{N/m}$
Right Segment ($L_2 = 0.1$ m): $k_2 = \frac{k L}{L_2} = 100 \times \frac{0.5}{0.1} = 500 \, \text{N/m}$

Step 2: Determine Effective Stiffness

When point P is pulled horizontally, the left segment is stretched and the right segment is compressed. Both segments resist the motion, so they act in parallel. $$ k_{eff} = k_1 + k_2 $$ $$ k_{eff} = 125 + 500 = 625 \, \text{N/m} $$

Step 3: Calculate Displacement

Using Hooke’s Law for the effective system: $$ x = \frac{F}{k_{eff}} $$ $$ x = \frac{5.00 \, \text{N}}{625 \, \text{N/m}} = 0.008 \, \text{m} $$ Converting to millimeters: $$ x = 0.008 \, \text{m} \times 1000 = 8.00 \, \text{mm} $$

Correct Option: (a) 8.00 mm