Question

A spring has natural length 28 cm. Compare the work W1 done in stretching the spring from 28 cm to 38 cm with the work W2 done in stretching it from 38 to 48 cm. (Use k for the spring constant)

Answer 1

Answer:

The work W₁ is 1/3 of the work W₂.

Explanation:

Due to the conservationof the mechanical energy, the work done is equal to the difference in mechanical energy. Assuming that when we stretch the spring its velocity reaches cero, the total mechanical energy is equal to the elastic potential energy of the spring. So:

$W_1=\Delta U_1=\frac{1}{2} kx_0^{2} -\frac{1}{2} kx_1^{2} =\frac{1}{2} k(0cm)^{2} -\frac{1}{2} k(10cm)^{2}=-50cm^{2} k\\\\W_2=\Delta U_2=\frac{1}{2} kx_1^{2} -\frac{1}{2} kx_2^{2} =\frac{1}{2} k(10cm)^{2} -\frac{1}{2} k(20cm)^{2}=-150cm^{2} k$

Finally, if we divide W₁/W₂, we obtain that W₁=1/3W₂