What is the speed of a block pulled by a spring with friction?

AI Thread Summary
A block with a mass of 1.50 kg is attached to a spring with a force constant of 456.0 N/m and is pulled 3.60 cm before being released. The coefficient of kinetic friction is 0.210, and the work done by friction is crucial for calculating the block's speed at equilibrium. The initial energy is zero since the block starts from rest, while the final energy includes kinetic energy minus the spring's potential energy. After correcting the friction force and accounting for its negative work, the correct speed of the block as it passes the equilibrium position is determined to be 0.496 m/s. Proper attention to signs and calculations is essential in solving such physics problems.
alexpratt
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Homework Statement



A block with mass 1.50 kg is attached as shown to a spring with a force constant of 456.0 N/m. The coefficient of kinetic friction between the block and the surface on which it slides is 0.210. The block is pulled 3.60 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?

Homework Equations



W=Ef-Ei

The Attempt at a Solution



W is the work done by friction, which would be f*d, where f is coefficient of friction and d is the distance the block is pulled back.
Ef is the final energy which would be the kinetic energy of the block minus the force of the spring since the spring is pushing on the block, however since I am looking at the equilibrium as my final position I am assuming this is 0?
Ei is the initial energy which is the kinetic energy minus the energy of the spring, but there's no velocity so kinetic energy is 0.


so i have my equation as

f*d=Ke(final)-Se(initial)
f*d=1/2mv^2 - 1/2kd^2

when i solve for v i get .636m/s which is incorrect.
i goofed up the equation somewhere, can anyone help me?
 
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alexpratt said:

Homework Statement



A block with mass 1.50 kg is attached as shown to a spring with a force constant of 456.0 N/m. The coefficient of kinetic friction between the block and the surface on which it slides is 0.210. The block is pulled 3.60 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?

Homework Equations



W=Ef-Ei

The Attempt at a Solution



W is the work done by friction, which would be f*d, where f is coefficient of friction and d is the distance the block is pulled back.
The work done by friction is (force of friction) (d), where the friction force is not the coefficient of friction. Friction force = ____?
Ef is the final energy which would be the kinetic energy of the block minus the force of the spring since the spring is pushing on the block, however since I am looking at the equilibrium as my final position I am assuming this is 0?
Ei is the initial energy which is the kinetic energy minus the energy of the spring, but there's no velocity so kinetic energy is 0.


so i have my equation as

f*d=Ke(final)-Se(initial)
f*d=1/2mv^2 - 1/2kd^2

when i solve for v i get .636m/s which is incorrect.
i goofed up the equation somewhere, can anyone help me?
The equation is OK, except correct the value of 'f'.
 
ok, so its f*mg*d? when i do that i get 0.737, last time i checked anyways. So maybe my math is off.
 
The work done by friction is negative...the force is opposiote to the direction of displacement, hence negative...watch that minus sign...it bites every time, if given the chance.:wink:
 
Last edited:
ahh, thank you! Got the answer to be 0.496 m/s, i almost ran out of tries too!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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