A horizontal spring is attached to a wall has a force constant of k=850 N/m. A b

Question

A horizontal spring is attached to a wall has a force constant of k=850 N/m. A block of mass m=1.00 kg is attached to the spring and rests on a frictionless, horizontal surface.

a.) The block is pulled to a position xi = 6.00 cm from the equilibrium position and released. Find the elastic potential energy stored in the spring when the block is 6.00 cm from equilibrium and when it passes through equilibrium.

b.) Find the speed of the block as it passes through the equilibrium point.

c.) What is the speed of the block when it is at a position xi 2 = 3.00 cm?

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 6cm all the energy is stored as potential in the spring.
Total energy = 1/2*k*x^2 = 1/2* 850 N/m * (.06 cm) ^2 = 1.53 J

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

1.53J = 1/2 * m * v^2
v = sqrt(2* 1.53 J / m ) = 1.7492 m/s

c) At xi/2 = 3.00 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*850 N/m * ( (.06 m)^2 - (.030 m)^2 ) = 1/2* 1 kg *vf^2

vf = sqrt(850 N / m / 1 kg * (.0036 m^2 - .00009 m^2) ) = 1.514 m/s

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