Question

A 0.5k block attached to a spring (k=200N/m) is sitting on top of a rough horizontal surface. Initially a spring is stretched 10 cm from its equillibrium position. When released, the block moves and stops 8 cm past the equillibrium position in the other direction. Find the friction coefficient between the block and surface.

Answer #1

^{2}

where K is spring constant = 200 N/m

and x is the initial amount of stretching = 10 cm = 0.1 m

= (1/2)*200*(0.1)^{2} = 1 J ---------------(1)

Final energy of the block = Potential energy of spring + Friction
energy

= (1/2)KX^{2} + W_{f}

Where X is compression of the spring = 8 cm = 0.08 m

W_{f} is the energy lost due to friction

Final energy = (1/2)*200*(0.08)^{2} +W_{f} = 0.64 +
W_{f} -------------(2)

applying energy conservation

Initial energy = Final energy

0.64 + W_{f} =1

W_{f} = 0.36 J

Now, We know that energy lost due to friction is given by

W_{f} = Friction Force*distance = (u*R_{N})*d

from the free body diagram

R_{N} = mg = 0.5*9.81 = 4.905 N and d = 18 cm = 0.18
m

W_{f} = u*(4.905)*0.18 = 0.883*u

u = W_{f}/0.883 = 0.36/0.883 = 0.41

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