A 1.30-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 1.30-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 15.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. (a)Find the speed of the object when its position is equal to one-third of the maximum value. (b) Find the acceleration of the object when its position is equal to one-third of the maximum value.

Explanation / Answer

This is an energy balance problem since there is no friction. Change in energy of the system = 0 Change in energy of the system = 1/2*k*(xf^2 - xi^2) + 1/2*m*(vf^2 - vi^2) a) When the spring is stretch/compressed by 5cm all the energy is stored as potential in the spring. Total energy = 1/2*k*x^2 = 1/2* 820 N/m * (.05 cm) ^2 = 1.025 J b) When it passes through the equilibrium point, all the spring energy is transferred into kinetic energy and the velocity is max 1.025J = 1/2 * m * v^2 v = sqrt(2* 1.025 J / m ) = 1.26 m/s c) At xi/2 = 2.50 cm you have part of the energy still in spring potential and part in kinetic energy Remember, the change in energy = 0 = 1/2*k*(xf^2 - xi^2) + 1/2*m*(vf^2 - vi^2) 1/2*k*(xi^2 - xf^2) = 1/2 * m * (vf^2 - vi^2) vi = 0 m/s 1/2*820 N/m * ( (.05 m)^2 - (.025 m)^2 ) = 1/2* 1.30 kg *vf^2 vf = sqrt(820 N / m / 1.30 kg * (.0025 m^2 - .000625 m^2) ) = 1.09 m/s

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