A block of mass m = 2.00 kg is attached to a spring of force constant k = 535 N/

Question

A block of mass m = 2.00 kg is attached to a spring of force constant k = 535 N/m as shown in the figure below. The block is pulled to a position xi = 4.60 cm to the right of equilibrium and released from rest.

(a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless.

            m/s

(b) Find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is ?k = 0.350.

               m/s

Explanation / Answer

a) we can use conservation of energy; the PE when the spring is extended equals the KE when the block passes through equilibrium

1/2 k x ^2 = 1/2 m v^2

v = Sqrt[ k x^2/m] = Sqrt[535N/m x 0.046m^2 / 2kg] = 0.75 m/s

b)

here, we have to take the work done by friction into account

we have in this case, initial PE = KE when block passes through origin + work done by friction

where the work done by friciton = friction force x distance = u m g x 0.046m

work by friction = 0.35x2kg x 9.8m/s/s x 0.046m = 0.315 J

therefore, the amount of energy available for the block's KE is

1/2 kx^2 - 0.315 J = 1/2 m v^2

(1/2 x 535 N/m x 0.046m^2 )- 0.315 J = 1/2 (2kg) v^2

v = 0.501 m/s

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