Given $k_{1}$ = 1500 N/m, $k_{2}$ = 500 N/m, $m_{1}$ = 2 kg, $m_{2}$ = 1 kg. Find work done in slowly pulling down $m_{2}$ by 8 cm.

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Solution

Here the springs are in series combination so,equivalent spring is constant,

K= $\frac{k_{1} k_{2}}{k_{i} + k_{2}}$

$\frac{1500 \times 500}{2000} = 375 N / m$

So the work done in slowly pulling down $m_{2}$ by 8 cm is,

W= $\frac{1}{2} K \times x^{2}$

= $\frac{1}{2} \times 375 \times 64 \times 10^{- 4}$

=1.2J

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Given k1 = 1500 N/m, k2 = 500 N/m, m1 = 2 kg, m2 = 1 kg. Find work done in slowly pulling down m2 by 8 cm.

Here the springs are in series combination so,equivalent spring is constant,K=k1k2ki+k21500×5002000=375N/mSo the work done in slowly pulling down m2 by 8 cm is,W=12K×x2=12×375×64×10−4=1.2J

Given $k_{1}$ = 1500 N/m, $k_{2}$ = 500 N/m, $m_{1}$ = 2 kg, $m_{2}$ = 1 kg. Find work done in slowly pulling down $m_{2}$ by 8 cm.