Question

17 mark for review note: figure not drawn to scale. a block is suspended from a single ideal spring. when the block is at equilibrium, the spring is stretched 0.10 m from its unstretched length. a second identical spring is then attached to the block as shown in the figure and the block is again allowed to come to equilibrium. what is the stretched length δy for the springs connected in parallel? a 0.025 m b 0.05 m c 0.10 m d 0.20 m

Explanation:

Step1: Apply Hooke's law for single - spring case

When the block is suspended from a single spring and at equilibrium, the gravitational force $F_g = mg$ is balanced by the spring force $F_{s1}=k\Delta y_1$. So $mg = k\times0.10$ where $\Delta y_1 = 0.10$ m and $k$ is the spring - constant of a single spring.

Step2: Consider two springs in parallel

When two identical springs of spring - constant $k$ are connected in parallel, the equivalent spring - constant $k_{eq}=k + k=2k$. At equilibrium, $mg=k_{eq}\Delta y$. Substitute $k_{eq}=2k$ and $mg = k\times0.10$ into the equation: $k\times0.10=2k\Delta y$.

Step3: Solve for $\Delta y$

Divide both sides of the equation $k\times0.10 = 2k\Delta y$ by $2k$ (since $k
eq0$). We get $\Delta y=\frac{0.10}{2}=0.05$ m.

Answer:

B. $0.05$ m