Question

a block hangs at rest at the end of an ideal spring with spring constant ( k_1 ) as shown in scenario 1. in scenario 2, a second ideal spring with a spring constant ( k_2 ), where ( k_2
eq k_1 ), is then added above the original spring and the block again hangs at rest. which statement best describes the equivalent spring constant ( k_{eq} ) in scenario 2?

a ( k_{eq}>k_1 ) because in scenario 2 both springs pull up on the block

b ( k_{eq}>k_1 ) because the combination of both springs will stretch more than the original spring

c ( k_{eq}<k_1 ) because the same weight will cause more stretching with two springs than with one

d ( k_{eq}<k_1 ) because in scenario 2 both springs stretch by the same amount.

Explanation:

Step1: Hooke's Law

Hooke's Law is \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the extension. In Scenario 1, \(F=k_1x_1\). In Scenario 2, the two springs are in series. For springs in series, \(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\).

Step2: Analyze the equivalent spring constant formula

Since \(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\) (\(k_2>0\)), then \(k_{eq}x_1\)) when there are two springs.

Answer:

C. \(k_{eq}