Elastic collision: Determine the deceleration

[thatstheguy9]

Homework Statement

A 8kg block with an initial velocity of 5 m/s travels 2 meters before colliding with a spring.
Spring constant is 200n/m and the coefficient of friction between the block and surface is 0.25.

I've calculated that the block compresses the spring 0.7m before it momentarily stops. The next part asks to determine a expression for the deacceleration of the block in terms of the spring compression, then use that to determine a expression for the velocity in terms of spring compression.

Relevant Equations

Conservation of energy
Friction Force = uN
F = ma

My approach so far is to use F = ma.

The forces acting on the block in the horizonital direction are friction and the force of the spring. Choosing the direction towards the spring as the positive axis.

Therefore: F = ma
-Fr - kx = ma

Solving for a = (-Fr - kx)/m

If I plug in values I end up with a = (-(0.25*9.81*8)-(200*0.25))/8
a = -25x-2.2452

However, when I plug 0.7 into the equation it does not resolve to zero. I am unsure where I'm going wrong here?

 

[PeroK]

However, when I plug 0.7 into the equation it does not resolve to zero. I am unsure where I'm going wrong here?

You still have the constant kinetic friction term in your equation. Technically, your equation is only valid while the block is moving (in the positive $x$ direction).

 

[haruspex]

If I plug in values I end up with a = (-(0.25*9.81*8)-(200*0.25))/8

I assume you mean a = (-(0.25*9.81*8)-(200*x)/8

when I plug 0.7 into the equation it does not resolve to zero

Why should it? You calculated that x value based on the velocity becoming zero, not the acceleration.

 

[kuruman]

However, when I plug 0.7 into the equation it does not resolve to zero. I am unsure where I'm going wrong here?

Does the deceleration (magnitude of the acceleration) increase, decrease or stay the same while the spring is compressed?

 

[bob012345]

Homework Statement:: A 8kg block with an initial velocity of 5 m/s travels 2 meters before colliding with a spring.
Spring constant is 200n/m and the coefficient of friction between the block and surface is 0.25.

I've calculated that the block compresses the spring 0.7m before it momentarily stops. The next part asks to determine a expression for the deacceleration of the block in terms of the spring compression, then use that to determine a expression for the velocity in terms of spring compression.

It would be helpful to know what the exact question as stated is. Thanks.

 

[thatstheguy9]

I assume you mean a = (-(0.25*9.81*8)-(200*x)/8

My apologies, yes that is correct.

Why should it? You calculated that x value based on the velocity becoming zero, not the acceleration.

I see. Thank you

 

[thatstheguy9]

It would be helpful to know what the exact question as stated is. Thanks.

 

[thatstheguy9]

You still have the constant kinetic friction term in your equation. Technically, your equation is only valid while the block is moving (in the positive $x$ direction).

I see. I'm unsure what to do then lol

 

[haruspex]

I see. I'm unsure what to do then lol

You obtained an expression for the acceleration in terms of the compression (i.e. displacement). Generically it takes the form:
$\ddot x+\alpha x+\beta=0$.
Do you recognise that? What kind of motion do you normally associate with a spring? What is the general form of the solution?

 

[PeroK]

I see. I'm unsure what to do then lol

If you want to look at the motion as the block rebounds, then you must change the direction of friction in your equation. But also, as noted by @haruspex, the block is always accelerating while being acted on by the spring - an instantaneous velocity of zero does not imply an acceleration of zero.

 

[thatstheguy9]

If you want to look at the motion as the block rebounds, then you must change the direction of friction in your equation. But also, as noted by @haruspex, the block is always accelerating while being acted on by the spring - an instantaneous velocity of zero does not imply an acceleration of zero.

My interpretation of the question was that we are looking at the motion as the block is slowed by the spring.
That makes sense, however I'm having struggling with deriving the velocity.

So far my acceleration = -25x - 2.452

To find the velocity I am using: a dx = v dv, is this the correct approach?

 

[thatstheguy9]

You obtained an expression for the acceleration in terms of the compression (i.e. displacement). Generically it takes the form:
$\ddot x+\alpha x+\beta=0$.
Do you recognise that? What kind of motion do you normally associate with a spring? What is the general form of the solution?

Sorry @haruspex I don't recognize that. Is the second derivative of displacement with respect to time? I haven't done any form of harmonic motion so the only concepts I have to solve it are particle dynamics and conservation of energy

 

[PeroK]

My interpretation of the question was that we are looking at the motion as the block is slowed by the spring.
That makes sense, however I'm having struggling with deriving the velocity.

So far my acceleration = -25x - 2.452

To find the velocity I am using: a dx = v dv, is this the correct approach?

You need to split the motion into two phases: first, sliding towards the spring; second, compressing the spring.

If you want velocity as a function of distance, then I would use energy.

Note that the diagram you posted shows the units associated with each quantity. You should do this too.

 

[thatstheguy9]

Thanks for the response @PeroK. I have already solved parts 1, 2 and 3 and used conservation of energy in two phases like you suggested for part 3. What's thrown me off the part 4. Are you saying I should apply $F = ma$ in two phases for this question?

Using energy to find the velocity as a function of distance would be my go-to also. However as the question states derive the velocity from the acceleration expression, the only way I could think to do so was to use $a ~ dx = v ~ dv$ and integrate both sides

 

[bob012345]

Thanks for the response @PeroK. I have already solved parts 1, 2 and 3 and used conservation of energy in two phases like you suggested for part 3. What's thrown me off the part 4. Are you saying I should apply $F = ma$ in two phases for this question?

Using energy to find the velocity as a function of distance would be my go-to also. However as the question states derive the velocity from the acceleration expression, the only way I could think to do so was to use $a ~ dx = v ~ dv$ and integrate both sides

Part #4 says to derive acceleration as a function of position $x$ but to then use that to develop an expression for velocity as a function of position $x$. I do not think it literally means you have to derive it from the acceleration formula you got but to just use it in another relation that relates the quantities you want.

 

[bob012345]

What was you speed of the block just as it gets to the spring?

 

[thatstheguy9]

What was you speed of the block just as it gets to the spring?

I got 3.9 m/s

 

[bob012345]

I got 3.9 m/s

Now, how far did the spring compress before the block stopped momentarily? Remember comment #2 by @PeroK .

 

[haruspex]

You need to split the motion into two phases: first, sliding towards the spring; second, compressing the spring.

What was you speed of the block just as it gets to the spring?

It isn't necessary to split it like that to answer the first part. A single energy equation can be applied from the initial given state until the spring reaches maximum compression:
$(2+x)\mu mg+\frac 12kx^2=\frac 12mv^2$.
In post #1, @thatstheguy9 gave 0.7m for the maximum compression, and to one more decimal place I get 0.69m.

That said, solving the second part using $a.dx=v.dv$ requires the acceleration function to be smooth. So one way is to split it into those two phases; another is only solve it for the compression phase but use the maximum displacement and terminal velocity zero as the boundary condition, avoiding the need to find the velocity on reaching the spring.

@thatstheguy9 , can you solve $a.dx=v.dv$ for the compression phase?

 

[PeroK]

Thanks for the response @PeroK. I have already solved parts 1, 2 and 3 and used conservation of energy in two phases like you suggested for part 3. What's thrown me off the part 4. Are you saying I should apply $F = ma$ in two phases for this question?

Using energy to find the velocity as a function of distance would be my go-to also. However as the question states derive the velocity from the acceleration expression, the only way I could think to do so was to use $a ~ dx = v ~ dv$ and integrate both sides

Yes, although if we tinker with that equation, we have $$Fdx = ma ~ dx = mv ~ dv$$And integrating gives the energy equation.