Question

one end of a spring of length l and spring constant k = 100 n/m² is attached to the ceiling as shown. the other end of the spring is attached to a small box of mass m_b = 0.6 kg, and the block is released from rest. the potential energy u_s of the box - spring system as a function of the stretch of the spring, δy, can be described as u_s = 1/2 k(δy)². the box is allowed to oscillate until the box - spring system comes to rest. the stretch of the spring, δy, when the box - spring system has come completely to rest is most nearly
a 0.06 m
b 0.2 m
c 0.5 m
d 0.8 m

Explanation:

Step1: Analyze the forces at equilibrium

When the system comes to rest, the force due to gravity \(F_g = m_bg\) (where \(g = 10m/s^{2}\)) is equal to the force exerted by the spring \(F_s=\frac{dU_s}{d(\Delta y)}\).
First, find \(F_s\):
Given \(U_s=\frac{1}{2}k(\Delta y)^{3}\), then \(F_s=\frac{dU_s}{d(\Delta y)}=\frac{3}{2}k(\Delta y)^{2}\) (using the power - rule for differentiation \(\frac{d}{dx}(x^{n})=nx^{n - 1}\)).
And \(F_g=m_bg\), with \(m_b = 0.6kg\) and \(g = 10m/s^{2}\), so \(F_g=0.6\times10=6N\).

Step2: Set up the equation and solve for \(\Delta y\)

Set \(F_s = F_g\), so \(\frac{3}{2}k(\Delta y)^{2}=m_bg\).
Substitute \(k = 100N/m\) and \(m_bg = 6N\) into the equation:
\(\frac{3}{2}\times100\times(\Delta y)^{2}=6\).
First, simplify the left - hand side: \(150(\Delta y)^{2}=6\).
Then, solve for \((\Delta y)^{2}\): \((\Delta y)^{2}=\frac{6}{150}=0.04\).
Take the square root of both sides: \(\Delta y=\sqrt{0.04}=0.2m\).

Answer:

B. \(0.2m\)