Question

A 0.75-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 6.7 to 3.2 m/s. What is the spring constant of the spring?

Answer 1

Question #37220

A 0.75-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 6.7 to 3.2 m/s. What is the spring constant of the spring?

Solution

Let

$$m = 0.75 \, kg$$ $$S_1 = 0.12 \, m$$ $$S_2 = 0.23 \, m$$ $$v_1 = 6.7 \, \text{m/s}$$ $$v_2 = 3.2 \, \text{m/s}$$ $$k = ?$$

According to the law of conservation energy

The change of the kinetic energy of sphere is equal to the change of potential energy of the spring

$$\Delta E_k = \Delta E_p$$ $$\Delta E_k = \frac{1}{2} m (v_1 - v_2)^2$$ $$\Delta E_p = \frac{1}{2} k (S_2 - S_1)^2$$

Following this

$$k = m \frac{(v_1 - v_2)^2}{(S_2 - S_1)^2}$$ $$k = 0.75 \frac{(6.7 - 3.2)^2}{(0.23 - 0.12)^2} = 760 \, \text{N/m}$$

Answer 760 N/m.