What Determines the Velocity of a Mass Released from a Compressed Spring?

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SUMMARY

The velocity of a mass released from a compressed spring can be calculated using the principles of energy conservation and kinematics. In this discussion, a 1.118 kg block with a spring constant of 715 N/m is analyzed, where the spring does 4.326 J of work upon release. The distance the spring was compressed is determined to be 0.11 m. To find the velocity, one must account for the work done against friction, applying the equation v = d/t, where t can be derived from t = √(2m/k).

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songokou77
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How do I figure velocity?!

What is the velocity of the mass as it breaks contact with the spring?
From part 1 I got:
A 1.118 kg block is on a horizontal surface with mu-k = 0.150, and is in contact with a lightweight spring with a spring constant of 715 N/m which is compressed. Upon release, the spring does 4.326 J of work while returning to its equilibrium position. Calculate the distance the spring was compressed. Answer:0.11 m
Also I got a hint:The force that acted on the mass comes from the spring and from friction. Thus the amount of work equals the kinetic energy of the spring. From the equation of work done by both forces we can calculate the velocity of the spring. (i thought i understood it, but something seems wrong.)

I thought of using the formula K=1/2(m*v^2) but it's wrong
Also from the hint o found that the force is: 39.33 N ( i divided the work done with the distance)
I also tried V^2=2*m*g*h, but i also get it wrong
What am i missing
 
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Your approach is partly correct. You need to know how to apply the law of conservation of energy to the system when friction forces are applicable, i.e. if you consider the block+spring to be your system, you should be able to classify the external forces on it.

As a hint, the friction force does work to oppose the motion of the block through the distance the spring tries to move it. So you should get something like -(1/2)kx^2 + (1/2)mv^2 = -fx where f is the frictional force. More generally if W_{ext} is the work done by a force other than gravitational or elastic (such as nonconservative forces like friction or drag), then the law can be written as

[tex]\Delta T + \Delta V_{g} + \Delta V_{e} = W_{ext}[/tex]

this is the same as saying that

Final KE + Final GPE + Final Elastic PE = Work done by friction + Initial KE + Initial GPE + Initial Elastic PE

where T = Kinetic Energy, V_{g} = Gravitational Potential Energy, V_{e} = Elastic potential energy.

Hope that helps...

Cheers
Vivek
 
?

To figure out velocity, you need to use the equation v = d/t, where v is velocity, d is distance, and t is time. In this case, you have the distance (the amount the spring was compressed) and you need to calculate the time it took for the spring to return to its equilibrium position.

To find the time, you can use the equation t = √(2m/k), where m is the mass of the block and k is the spring constant. Plug in the values given in the problem to solve for t.

Once you have the time, you can plug it into the equation v = d/t to calculate the velocity of the mass as it breaks contact with the spring.

Remember to always double check your units and make sure they are consistent throughout the calculation. If you are still having trouble, it may be helpful to draw a diagram and label all the given values to better visualize the problem.
 

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