A block with mass 2.30 kg is attached as shown to a spring with a force constant of 418.0 N/m. The coefficient of kinetic friction between the block and the surface on which it slides is 0.240. The block is pulled 6.90 cm to the right of its equilibrium position and then released from rest. What is the speed of the block as it passes by its equilibrium position?
Refer to this image link:
http://loncapa.physics.mun.ca/res/mun/PHYSICS/munphysicslib/Graphics/005_Work%20and%20Energy/horiz_spring.gif
My gratitude to your time.
Refer to this image link:
http://loncapa.physics.mun.ca/res/mun/PHYSICS/munphysicslib/Graphics/005_Work%20and%20Energy/horiz_spring.gif
My gratitude to your time.
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When the spring is stretched from its equilibrium position, it has a potential energy of PE = ½*k*x²
The system energy when the block passes its equilibrium position is the kinetic energy of the block
KE = ½*m*v²
This kinetic energy must equal the initial PE less the energy lost to friction, Ef = friction force times distance moved = µ*m*g*x
½*m*v² = ½*k*x² - µ*m*g*x
v = √[(k/m)*x² - 2*µ*g*x]
k = 418.0 N/m
m = 2.30 kg
x = 0.0690 m
µ = 0.240
v = 0.735 m/s
The system energy when the block passes its equilibrium position is the kinetic energy of the block
KE = ½*m*v²
This kinetic energy must equal the initial PE less the energy lost to friction, Ef = friction force times distance moved = µ*m*g*x
½*m*v² = ½*k*x² - µ*m*g*x
v = √[(k/m)*x² - 2*µ*g*x]
k = 418.0 N/m
m = 2.30 kg
x = 0.0690 m
µ = 0.240
v = 0.735 m/s