A horizontal spring attached to a wall has a force constant of k = 820 N/m.

A horizontal spring attached to a wall has a force constant of k = 820 N/m. A block of mass m = 1.30 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below.
(a) The block is pulled to a position xi = 5.00 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.00 cm from equilibrium. = 1.025J

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



Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

(c) What is the speed of the block when it is at a position xi/2 = 2.50 cm?
Your response differs...
so confused help on last 2 please

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