A horizontal spring attached to a wall has a force constant of 810 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 810 N/m. A block of mass 1.10 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released.

(a) What objects constitute the system, and through what forces do they interact?

(b) What are the two points of interest?

(c) Find the energy stored in the spring when the mass is stretched 6.60 cm from equilibrium and again when the mass passes through equilibrium after being released from rest.

x = 6.60 cm

x = 0 cm

(d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value.

(e) What is the speed at the halfway point?

Why isn't it half the speed at equilibrium?

x = 6.60 cm

x = 0 cm

The spring force always acts toward the equilibrium point, which is at x= 0 in this figure. x=0 For an equilibrium point at x= 0, the spring potential energy is RX. KEi = 0 x=0 PE,-0 f= mu

Explanation / Answer

force constant k = 810 N/m

mass of the block m = 1.10 kg

a) spring, wall and mass of the object constitute the system.

the gravitational force , restoring force of spring and normal (reaction) force are interact.

b) point of interest : equilibrium and extreme position

c) at equilibrium, displacement = 0 , so , kinetic energy = 0 J

therefore,energy stored in the spring is

         U = 1/2 kx2 = (1/2)(810)(6.6 x 10-2)2 = 1.764 J

d) from law of conservation of energy,

     Ki + Ui = Kf + Uf

     (1/2)mvi2 + (1/2) kxi2 = (1/2) mvf2 + (1/2) kxf2

      mvi2 + kxi2 = mvf2 + kxf2

     speed vf = [vi2 + (k/m)(xi2 - xf2)]1/2     ................ (1)

substitute the given data in above eq , we get

                 vf = [(0)2 + (810/1.1)((0.066)2 - (0)2)]1/2

                 vf = 1.79 m/s

e) from eq (1),

         speed vf = [vi2 + (k/m)(xi2 - xf2)]1/2

where , vi = 0 m/s , xi = 0.066 m and xf = 0.066/2 = 0.033 m

          speed vf = [(0)2 + (810/1.1)((0.066)2 - (0.033)2)]1/2

          speed vf = 1.55 m/s

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