Calculating Speed of Block-Spring Collision w/Kinetic Friction

AI Thread Summary
The discussion focuses on calculating the speed of a block at the moment it collides with a spring on a horizontal surface, factoring in kinetic friction. The initial equation relating elastic potential energy and kinetic energy is modified to include the work done against friction. The work done by friction is expressed as the product of the frictional force and the distance moved. The correct energy conservation equation incorporates the energy lost to friction, leading to the formulation of the speed in terms of various parameters. Understanding the role of the coefficient of kinetic friction is crucial for accurate calculations in this scenario.
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one more spring question :smile:


A block collides with a spring on a horizontal surface

the block compresses the spring x metres from rest position. What was its speed at the instant of collision

Ok so far I know how to do this question

Ee = Ek
.5kx^2 = .5mv^2


but in this question there is a coefficient of kinetic friction between the block and the surface.. not sure how it fits in.

I'm thinking that it is only affecting the Ee part of the equation..so:

Ee = Ek
.5ukx^2 = .5mv^2

is this right?
 
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Well, you can still use conservation of energy, only you have to take into account the work done against friction as well. The friction is a constant force, so things are kept simple.

Do you know F_{fr} = \mu N, to relate the frictional force to the normal force of the surface on the block ? N, is of course, equal to the weight of the block considering that this is a horizontal surface.

The work done against friction is (F_{fr})(x), since x is the distance moved in the direction of the force.

Then just use conservation of energy :

Initial energy of system = Final energy of system

Initial KE of block = Work done against friction (dissipated as heat to surface and block) + Potential Energy stored in spring.

\frac{1}{2}mv^2 = \mu mgx + \frac{1}{2}kx^2

and express v in terms of \mu, m, g, k and x.
 
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Yes, you are correct in thinking that the coefficient of kinetic friction will affect the calculation. In this case, the energy lost due to friction will need to be taken into account in the equation. So the correct equation would be:

Ee = Ek + Ef

where Ee is the elastic potential energy, Ek is the kinetic energy, and Ef is the energy lost due to friction. So the equation would be:

.5kx^2 = .5mv^2 + uFn

where u is the coefficient of kinetic friction and Fn is the normal force. From here, you can solve for v to find the speed of the block at the instant of collision. Keep in mind that the normal force will also need to be calculated using the weight of the block and the angle of the surface. I hope this helps!
 
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