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Chapter 9: Q. 27 (page 228)

A $10$ -cm-long spring is attached to the ceiling. When a $2.0$ kg mass is hung from it, the spring stretches to a length of $15$ cm.
a. What is the spring constant?
b. How long is the spring when a 3.0 kg mass is suspended from it?

Short Answer

Expert verified

a). The spring constant is $392$ N/m

b). The spring when a $3.0$ kg mass it long $17.5$ cm

Step by step solution

01

Step 1. Given Information

Length of spring $L_{o} = 10 c m = 0.1 m = s_{e q}$

mass of load $m = 2$ Kg

Stretched length of spring $s = 15 c m = 0.15 m$

02

Step 2. Part a). The spring constant K

In this case as the load hangs from the spring, it streches because of the weight of load. So the force acting on the spring is weight of load

From Hooke's law:

${(F_{s p})}_{s} = K 鈭� s K = \frac{{(F_{s p})}_{s}}{鈭� s} 鈭� s = s - s_{e q} {(F_{s p})}_{s} = m g = ((2 k g) 脳 (9.8 m / s^{2})) N = 19.6 N K = (\frac{19.6}{0.15 - 0.1}) N / m K = 392 N / m$

03

Step 3. Part b). To find the length of the spring when mass is increased to 3 Kg

As the spring is the same hence its spring constant will remain the same.

Again from Hooke's law

${(F_{s p})}_{s} = K 鈭� s 鈭� s = \frac{{(F_{s p})}_{s}}{K} {(F_{s p})}_{s} = m g = 3 脳 9.8 = 29.4 N 鈭� K = 392 N / m s - s_{e q} = \frac{29.4 N}{392 N / m} = 0.075 m s = 0.075 m + 0.1 m = 0.175 m = 17.5 c m$

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