Question

1. a student uses a force probe to exert varying forces on the ends of vertical springs a, b, and c. the student then sketches the magnitude of the applied force as a function of the length of the spring for each spring, as shown.

chart: force applied to spring (y - axis) vs. length of spring (x - axis), with lines a, b, c

which of the following correctly compares the spring constants ( k ) of each spring?

a. ( k_b > k_c > k_a )
b. ( k_b > k_a > k_c )
c. ( k_c > k_b > k_a )
d. ( k_a > k_b > k_c )

Explanation:

Step1: Recall Hooke's Law

Hooke's Law states that the force \( F \) applied to a spring is related to its displacement \( x \) (change in length) by \( F = kx \), where \( k \) is the spring constant. In a graph of \( F \) vs. spring length (where the x - axis can be thought of as including the original length and the stretched length, so the slope of the \( F \)-length graph is related to \( k \)). The steeper the slope of the \( F \)-length graph, the larger the spring constant \( k \), because \( F = k(\Delta L) \), and if we consider the graph of \( F \) vs. \( L \) (where \( L = L_0+\Delta L \), \( L_0 \) is the original length), the slope \( m=\frac{\Delta F}{\Delta L}=k \).

Step2: Analyze the slopes of the graphs

Looking at the graphs of springs A, B, and C:

• Spring A has the steepest slope among the three.
• Spring B has a slope less steep than A but steeper than C.
• Spring C has the least steep slope.

Since the slope of the \( F \)-length graph is equal to the spring constant \( k \) (from \( F = k\Delta L \) and \( \Delta L=L - L_0 \), so \( F=k(L - L_0)=kL - kL_0 \), the slope of \( F \) vs. \( L \) is \( k \)), a steeper slope means a larger \( k \). So the order of spring constants is \( k_A>k_B>k_C \).

Answer:

d. \( k_A > k_B > k_C \)