Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.068 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 17.8 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.034 m relative to its unstrained length

Answer 1

Answer:

v_f = 1.05 m/s

Explanation:

From conservation of energy;

E_f = E_i

Thus,

(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²

This reduces to;

(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²

Making v_f the subject, we have;

v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]

We know that ω = √(k/m)

Thus,

v_f = ω[√((x_i)² - (x_f)²)]

Plugging in the relevant values to obtain;

v_f = 17.8[√((0.068)² - (0.034)²)]

v_f = 17.8[0.059] = 1.05 m/s