Question

1. a spring hangs at rest from a support. if you suspend a 460 g mass from the spring, its deflection is 7.9 cm shown in the system diagram to the right. determine the spring constant. a. 5.7 n/m b. 57 n/m c. 58 n/m d. 570 n/m

Explanation:

Step1: Convert units

Convert mass \(m = 460g=0.46kg\) and deflection \(\Delta x = 7.9cm = 0.079m\).

Step2: Apply Hooke's law

When the mass is in equilibrium, the force of gravity \(F_g=mg\) is equal to the spring force \(F_s = k\Delta x\). So \(k=\frac{mg}{\Delta x}\).
Substitute \(m = 0.46kg\), \(g = 9.8m/s^{2}\), and \(\Delta x=0.079m\) into the formula: \(k=\frac{0.46\times9.8}{0.079}\).
Calculate \(0.46\times9.8 = 4.508\), then \(k=\frac{4.508}{0.079}\approx57N/m\).

Answer:

B. \(57\ N/m\)