Question

A 1.3-kg mass is attached to a spring with a force constant of
52 N>m. If the mass is released with a speed of 0.28 m>s at a distance of 8.1 cm from the equilibrium position of the spring, what
is its speed when it is halfway to the equilibrium position?

Answer 1

Given info : A 1.3-kg mass is attached to a spring with a force constant of 52 N/m. If the mass is released with a speed of 0.28 m/s at a distance of 8.1 cm from the equilibrium position of the spring.

To find : the speed when it is halfway to the equilibrium position, is...

solution : initial velocity, u = 0.28 m/s

initial position, x₁ = 8.1 cm = 0.081 m

final position, x₂ = x₁/2 = 0.0405 m

using energy conservation theorem,

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

⇒1/2 mu² + 1/2 kx₁² = 1/2 mv² + 1/2 kx₂²

⇒m(v² - u²) = k(x₁² - x₂²)

⇒1.3(v² - 0.28²) = 52(0.081² - 0.0405²)

⇒1.3(v² - 0.0784) = 0.256

⇒v² - 0.0784 ≈ 0.2

⇒v² = 0.2784

⇒v ≈ 0.53 m/s

Therefore the speed of mass is 0.53 m/s