Solve Spring Equilibrium Problem: (a) & (b) w/ Friction

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Homework Help Overview

The problem involves a block attached to a spring, exploring its motion through the equilibrium position under different conditions, including the presence and absence of friction. The subject area includes mechanics, specifically dynamics and energy conservation principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, including kinetic and potential energy relationships. There are attempts to calculate the speed of the block at the equilibrium position under varying conditions of friction. Some participants question the correctness of unit usage and the sign of work done by friction.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the validity of their approaches. Some have provided insights into the work-energy theorem and its application to the problem, while others express confusion regarding specific calculations and concepts.

Contextual Notes

There is a mention of unit conversion issues, as the spring is stretched a distance of 0.33 cm, which may affect calculations. The problem also involves determining the effects of friction on the system's energy and motion.

cgotu2
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Please help with this problem!

A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.
(a) Calculate the speed of the block as it passes through the equilibrium position if no friction is present.

(b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 2.0 N retards its motion.

(c) What would be the strength of the frictional force if the block reached the equilibrium position the first time with zero velocity?

I got part (a) by by saying that 1/2mv^2=1/2kA^2 and my answer of 11.898 cm/s was correct. but i can't get (b) or (c). in (b) i said that the work done by friction (2*0.33) was equal to 1/2mv^2 - 1/2kA^2, but my answer is incorrect.

i have no clue how to do (c). please help me!
 
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for part b, the work done by friction is negative
 
okay i did that, and i got 11.86 cm/s, but it is still wrong (for b)
 
cgotu2 said:
Please help with this problem!

A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.

it's stretched 0.33 centimeters?

make sure you use the right units, i guess..
 
From the work-kinetic energy theorem we have that the work done by all of the forces on a system is equal to the change in i'ts kinetic energy or
W=\Delta K
We can split the work done into the two basis types for conservative an non conservative forces
W_{nc} \ + \ W_c=\Delta K
and since the work done by the conservative force is equal to the negative of the change in the sytem's potential energy or
W_c=-\Delta U
we can change the previous relation to
W_{nc} = \Delta K \ + \ \Delta U
In this case the non conservative force is friction.

The last question is actually quite easy if one can answer this question: How much energy must be removed from the system, by the frictional force, in order to bring the mass to a halt at the equilibrium position?
 
Last edited:

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