A horizontal spring attached to a wall has a force constant of k = 860 N/m. A bl

Question

A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.70 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 = 7.00 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 7.00 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 = 3.50 cm?
m/s

Explanation / Answer

Before the block is moved we have... weight of block acting vertically downward = mg = 9.8 N reaction of surface = 9.8 N coeffiecient of friction = 0.18 spring constant = 563 N/m By definition, the frictional force = coefficient of friction * reaction force = 0.18 * 9.8 = 1.76 N The force required to pull the block 7 cm (0.07 m) from equilibrium = 563 * 0.07 = 39.41 N But we also have to overcome the frictional force. So actual pulling force = 39.41 + 1.76 = 41.17 N The work done = force * distance = 41.17 * 0.07 = 2.88 J When the block passes through equilibrim, its potential energy will have been converted into kinetic energy. So KE = (1/2) m v^2 = 2.88 J v^2 = 2.88 * 2 / 1 = 5.76 v = v5.76 = 2.4 m/s

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