You attach one end of a spring with a force constant k = 773 N/m to a wall and t

Question

You attach one end of a spring with a force constant k = 773 N/m to a wall and the other end to a mass m = 2.12 kg and set the mass-spring system into oscillation on a horizontal frictionless surface as shown in the figure. To put the system into oscillation, you pull the block to a position xi = 6.56 cm from equilibrium and release it.

(a) Determine the potential energy stored in the spring before the block is released.
_________________J

(b) Determine the speed of the block as it passes through the equilibrium position.
_________________m/s

(c) Determine the speed of the block when it is at a position xi/4.

_________________m/s

Explanation / Answer

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* 773 N/m * (.0656 cm) ^2 = 1.6632J

b) When it passes through the equilibrium point, all the spring energy is transferred into kinetic energy and the velocity is max

1.6632J = 1/2 * m * v^2
v = sqrt(2* 1.6632 J / m ) = 1.252 m/s

c) At xi/4 = 1.64 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*773 N/m * ( (.0656m)^2 - (.0164 m)^2 ) = 1/2* 2.12 kg *vf^2

vf = sqrt(773 N / m / 2.12 kg * (.00430336 m^2 - .0002689 m^2) ) = 1.212 m/s

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