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A 2.00 kg block is attached to a spring of force constant 560 N/m as in Figure. The block is pulled 5.95 cm to the right of equilibrium and released from rest.

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

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

Ee = 0.5kx^2

F = -kx

Ek = 0.5mv^2

Ff = umg

A)

So Ee = 0.5 (560 N/m) (0.0595 m)^2

= 0.99127

Ee gets tranferred into Ek

So 0.99127 = 0.5 mv^2

1.98254 = (2 kg) v^2

v = 0.9956

B)

Ff = umg

= (0.350)(2 kg)(9.8)

= 6.86

I don't kno how to relate the frictional force into energy. I know that the block should be slower than in part A) tho.

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# Springs Help Needed Plz

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