Question

A horizontal spring attached to a wall has a force constant of k = 770 N/m. A block of mass m = 2.00 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 = 5.80 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.80 cm from equilibrium.

J

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

m/s

(c) What is the speed of the block when it is at a position xi/2 = 2.90 cm?

m/s

Answer 1

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 5.80 cm all the energy is stored as potential in the spring.
Total energy = 1/2*k*x^2 = 1/2* 770 N/m * (.058 cm) ^2 = 1.295 J

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

1.295J = 1/2 * m * v^2
v = sqrt(2* 1.295 J / m ) = 1.61 m/s

c) At xi/2 = 2.90 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*770 N/m * ( (.058 m)^2 - (.029 m)^2 ) = 1/2* 2.00 kg *vf^2

vf = sqrt(770 N / m / 2.00 kg * (.00336 m^2 - .000841 m^2) ) = 0.985 m/s

Hope It will help you.

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