Given $k_{1} = 1500 N m^{- 1}, k_{2} = 500 N m^{- 1}, m_{1} = 2 k g, m_{2} = 1 k g$ . Find:
a. Potential energy stored in the springs in equilibrium, and
b. work done in slowly pulling down $m_{2}$ by $8 c m$ .

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Solution

Let $x_{1} & x_{2}$ be the extensions in the spring with spring cosntant $k_{1} & k_{2}$ respectively.
(a) in equilibrium

$k_{1} x_{1} = (m_{1} + m_{2}) g$

$x_{1} = \frac{(m_{1} + m_{2}) g}{k_{1}} = \frac{(2 + 1) 10}{1500} = 0.02 m$

$k_{2} x_{2} = m_{2} g$

$x_{2} = \frac{m_{2} g}{k_{2}} = \frac{1 \times 10}{500} = 0.02 m$

Total potential energy, $P = \frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{2} x_{2}^{2}$

$P = \frac{1}{2} 1500. {(0.02)}^{2} + \frac{1}{2} 500. {(0.02)}^{2} = 0.4 J$

Hence, potential energy is $0.4 J$

(b) When pulling, spring in series pull together.

$k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{1500 \times 500}{1500 + 500} = 375 N m^{2}$

Potential energy, $P = \frac{1}{2} k x^{2} = \frac{1}{2} \times 375 \times {0.08}^{2} = 1.2 J$

Hence, Potential energy is $1.2 J$

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Given k1=1500 Nm−1,k2=500 Nm−1,m1=2kg,m2=1kg. Find:a. Potential energy stored in the springs in equilibrium, and b. work done in slowly pulling down m2 by 8 cm.

Given $k_{1} = 1500 N m^{- 1}, k_{2} = 500 N m^{- 1}, m_{1} = 2 k g, m_{2} = 1 k g$ . Find:
a. Potential energy stored in the springs in equilibrium, and
b. work done in slowly pulling down $m_{2}$ by $8 c m$ .