A block of mass 2.0 kg is attached to a horizontal spring that has a force const

Question

A block of mass 2.0 kg is attached to a horizontal spring that has a force constant of 2,500 N/m as shown in figure (a). The spring is compressed 2.5 cm and is then released from rest as in figure (b) (A) Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionless. x=0 (B) Calculate the speed of the block as it passes through the equilibrium position if a constant friction force of 5.0 N retards its motion from the moment it is released. SOLVE IT (A) Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionless x=0

Explanation / Answer

A] At the equilibrium position, the elastic potential energy = kinetic energy

(1/2)kx2 = (1/2)mv2

(1/2)(2500)(0.025)2 = (1/2)(2)v2

=> v = 0.884 m/s

B] Kinetic energy at the equilibrium position = (1/2)kx2 - F[2 x 0.025] = 0.53125 J

so, 0.53125 = (1/2)2v2

=> v = 0.7288 m/s

C] If the frictional force is F = 8.7 N now,

v = [(1/2)(2500)(0.025)2 - 8.7(2x0.025)]1/2 = 0.5884 m/s.

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