A horizontal spring attached to a wall has a force constant of k = 700 N/m. A bl

Question

A horizontal spring attached to a wall has a force constant of k = 700 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 = 5.60 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.60 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 = 2.80 cm?
m/s

A horizontal spring attached to a wall has a force constant of k = 700 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 = 5.60 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.60 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 = 2.80 cm? m/s

Explanation / Answer

Total energy = 1/2*k*x^2 = 1/2* 700 N/m * (.056 cm) ^2 = 1.0976 J

b)1.0976J = 1/2 * m * v^2
v = sqrt(2* 1.0976 J / 1.3 ) = 1.299 m/s

c)c) At xi/2 = 2.80 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*700 N/m * ( (.056 m)^2 - (.028 m)^2 ) = 1/2* 1.30 kg *vf^2

vf = sqrt(700 N / m / 1.30 kg * (.003136 m^2 - .000784 m^2) ) = 478 m/s

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