k A block of mass m = 2.00 kg is attached to a spring of force constant k = 495 N/m as shown in the figure below. The block is pulled to a position x, = 4.40 cm to the right of equilibrium and released from rest. x=0 Submit Answer 771 x = x₁ O (a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless. 0.692 ✓ m/s Read It TO 8.5.01.008. MY NOTES (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 = 0.350. x The force of friction causes an increase in the internal energy of the system and a decrease in the total mechanical energy. m/s Need Help? Watch It ASK YOUR TEACHER

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A block of mass m = 2.00 kg is attached to a spring of force constant k = 495 N/m as shown in the figure below. The block is pulled to a position x, = 4.40 cm to the right of
equilibrium and released from rest.
k
x=0)
O
(a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless.
0.692
✔ m/s
Submit Answer
777
MY NOTES
(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 μ = 0.350.
The force of friction causes an increase in the internal energy of the system and a decrease in the total mechanical energy. m/s
Need Help?
Read It
ASK YOUR TEACHER
Watch It
Transcribed Image Text:A block of mass m = 2.00 kg is attached to a spring of force constant k = 495 N/m as shown in the figure below. The block is pulled to a position x, = 4.40 cm to the right of equilibrium and released from rest. k x=0) O (a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless. 0.692 ✔ m/s Submit Answer 777 MY NOTES (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 μ = 0.350. The force of friction causes an increase in the internal energy of the system and a decrease in the total mechanical energy. m/s Need Help? Read It ASK YOUR TEACHER Watch It
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