Question

A 0.51-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 3.3 m/s. What is the spring constant of the spring

Answer 1

Answer:

487.23 N/m

Explanation:

Given:

mass of metal sphere 'm'= 0.51kg

the spring stretches from 0.12 m to 0.23 m. Therefore,

$s_1}$= 0.12m and $s_{2}$= 0.23m

the speed of the sphere decreases from 6.7 to 3.3 m/s. Therefore,

$v_{1}$= 6.7m/s and $v_2}$=3.3m/s

In order to find spring constant, we apply law of conservation of energy. i.e

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

So, Δ$E_{k} =$ Δ$E_{v}$

where,

Δ$E_{k} =$ 1/2 m ($v_{1}$- $v_2}$)²

Δ$E_{v}$= 1/2 k ($s_1}$ - $s_{2}$)²

1/2 m ($v_{1}$- $v_2}$)² = 1/2 k ($s_{2}$- $s_1}$ )²

k= m [ ($v_{1}$- $v_2}$)²/($s_{2}$- $s_1}$ )²

k= 0.51 [(6.7-3.3)²/ (0.23-0.12)²]

k= 487.23 N/m

Thus, the spring constant of the spring is 487.23 N/m