Question

A 53.6 g mass is attached to a horizontal spring with a spring constant of 8.84 N/m and released from rest with an amplitude of 21.3 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.

Answer 1

The speed of the mass halfway to the equilibrium position is 0.55 m/s.

Step-by-step explanation:

if K is the kinetic energy and P is the potential energy then the total energy of the system is:

Etot = K + P

we given the amplitude, and the halfway distance is x = 1/2×A them we can get Pmax by:

Pmax = P + K

1/2×k×A^2 = 1/2×k×x^2 + 1/2×m×v^2

k×A^2 = k×x^2 + m×v^2

m×v^2 = k×(A^2 - x^2)

v^2 = k×(A^2 - x^2)/m

= (8.84)×((21.3×10^-2)^2 - (1/2(21.3×10^-2)^2))

= 0.30079647

v = 0.55 m/s

Therefore, the speed of the mass halfway to the equilibrium position is 0.55 m/s.