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 0.065 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 14.7 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.032 m relative to its unstrained length

Answer 1

0.832 m/s

Step-by-step explanation:

The work done by the spring W equals the kinetic energy of the object K

The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀ = initial compression = 0.065 m and x₁ = final compression = 0.032 m

The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object

Since W = K,

1/2k(x₀² - x₁²) = 1/2mv²

k(x₀² - x₁²) = mv²

mv² = k(x₀² - x₁²)

v² = [(k/m)(x₀² - x₁²)]

taking square root of both sides, we have

v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),

v = √[(k/m)√(x₀² - x₁²)]

v = ω√(x₀² - x₁²)]

Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have

v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]

v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]

v = 14.7 rad/s × √(0.003201 m²)

v = 14.7 rad/s × 0.056577

v = 0.832 m/s