Dynamics of a Block on top of a slab with friction between them

[Thermofox]

Homework Statement

A block, with mass $m_b = 1 kg$, is placed on a slab, with mass $m_s= 4 kg$, which is able to slide without friction on the horizontal plane on which it is placed, as shown in the Figure. There is friction between the body and the slab characterized by a coefficient of dynamic friction $μ_d=0.2$. The body is initially placed at the left end of an ideal spring, with elastic constant $k = 600 N/m$, whose right end is fixed to the right end of the slab. The spring, initially compressed by $Δl = 5 cm$, at a certain instant is released to expand freely. Assuming the bodies are initially at rest and knowing that the distance separating the body of mass m from the left edge of the slab is $l = 20 cm$

Relevant Equations

$\Sigma F = ma$

I need to determine:
1) The accelerations of both the slab and the block, the moment right after the spring was released.
=> I can consider Fspring as a constant force. Both bodies can be considered as points of mass.
I'm taking as positive the left direction. I've analyzed both objects using the free body diagram and found that:
$$\begin{cases}
F_{spring}-f_d=m_b a_b \\
f_d= m_s a_s \\
f_d= \mu_d m_b g \\
\Sigma F_y=0
\end{cases}$$

From this system I can determine both accelerations since I know everything, except them.

$$\begin{cases}
k \Delta l - \mu_d m_b g = m_b a_b \\
m_b \mu_d g= m_s a_s
\end{cases}$$

$\begin{cases} a_b= \frac {k \Delta l - m_b g \mu_d} {m_b} = \frac {(600) (0.05) - (1)(9.81)(0.2)} 1 = 28.0 \frac m {s^2} \\ a_s= \frac {m_b g \mu_d} {m_s}= \frac {(1)(9.81)(0.2)} {4}= 0.49 \frac m {s^2} \end{cases}$

2)The velocities of both the block and the slab, when the block reaches the left edge of the slab.

Here I've tried using an energy balance equation:
$ΔE_m=W_{\text{non conservative forces}}$
=> $\frac 1 2 m_s v_{s,f}^2 + \frac 1 2 m_b v_{b,f}^2 - \frac 1 2 k \Delta l^2 = - W_{\text{friction}}$
But I don't know how to go forward, since I have 2 unknowns $(v_{b,f};v_{s,f})$ but only one equation.
I thought of maybe finding $v_{s,f}$ by observing that the slab is under the action of a constant force $f_d$ (since $f_d$ depends only by the mass of the block). Therefore the slab moves in a uniformly accelerated motion, but I neither have the final time nor the final position.

 

[BvU]

Homework Statement: A block, with mass $m_b = 1$ kg, is placed on a slab, with mass $m_s= 4$ kg, which is able to slide without friction on the horizontal plane on which it is placed, as shown in the Figure.

You could post the figure ...

There is friction between the body and the slab characterized by a coefficient of dynamic friction $μ_d=0.2$. The body is initially placed at the left end of an ideal spring, with elastic constant $k = 600$ N/m, whose right end is fixed to the right end of the slab.

So: does the spring exert a force on the slab?

The spring, initially compressed by $Δl = 5$ cm, at a certain instant is released to expand freely. Assuming the bodies are initially at rest and knowing that the distance separating the body of mass m from the left edge of the slab is $l = 20$ cm

Relevant Equations: $\Sigma F = ma$

I need to determine:
1) The accelerations of both the slab and the block, the moment right after the spring was released.
=> I can consider Fspring as a constant force. Both bodies can be considered as points of mass.
I'm taking as positive the left direction. I've analyzed both objects using the free body diagram and found that:
$$\begin{cases}
F_{spring}-f_d=m_b a_b \\
f_d= m_s a_s \\
f_d= \mu_d m_b g \\
\Sigma F_y=0
\end{cases}$$

Please post your free body diagram(s?)

From this system I can determine both accelerations since I know everything, except them.

$$\begin{cases}
k \Delta l - \mu_d m_b g = m_b a_b \\
m_b \mu_d g= m_s a_s
\end{cases}$$

$\begin{cases} a_b= \frac {k \Delta l - m_b g \mu_d} {m_b} = \frac {(600) (0.05) - (1)(9.81)(0.2)} 1 = 28.0 \frac m {s^2} \\ a_s= \frac {m_b g \mu_d} {m_s}= \frac {(1)(9.81)(0.2)} {4}= 0.49 \frac m {s^2} \end{cases}$

Are you saying block and slab are moving in the same direction ?

$\$

 

[Thermofox]

You could post the figure ...

So: does the spring exert a force on the slab?


Please post your free body diagram(s?)


Are you saying block and slab are moving in the same direction ?

$\$

I forgot to post them, sorry

 

[Thermofox]

Are you saying block and slab are moving in the same direction ?

It also seems odd to me, but I can't understand why.

 

[BvU]

does the spring exert a force on the slab?

 

[Thermofox]

So: does the spring exert a force on the slab?

$a_s= \frac {m_b g \mu_d -(k \Delta l)} {m_s}= \frac {(1)(9.81)(0.2)- (600)(0.05)} {4}= -7.00 \frac m {s^2}$
I didn't think that the spring was also connected to the slab.

 

[haruspex]

I didn't think that the spring was also connected to the slab.

"whose right end is fixed to the right end of the slab"

 

[Thermofox]

"whose right end is fixed to the right end of the slab"

Say it to my mind....
What could I do to find the velocities? I need another equation but can't find it.

 

[Thermofox]

Since the spring is what causes the motion can I split and analyze, separately, this 2 systems:
1)The spring and the block that moves onto a rough plane
2)The spring and the slab that moves onto a rough plane
then I would have:

1) $\frac 1 2 m_b v_{b,f}^2 - \frac 1 2 k \Delta l^2 = -W_f$
2)$\frac 1 2 m_s v_{s,f}^2 - \frac 1 2 k \Delta l^2 = -W_f$

 

[haruspex]

Since the spring is what causes the motion can I split and analyze, separately, this 2 systems:
1)The spring and the block that moves onto a rough plane
2)The spring and the slab that moves onto a rough plane
then I would have:

1) $\frac 1 2 m_b v_{b,f}^2 - \frac 1 2 k \Delta l^2 = -W_f$
2)$\frac 1 2 m_s v_{s,f}^2 - \frac 1 2 k \Delta l^2 = -W_f$

Be careful not to double count the work done against friction or by the spring.
What are you asked to find? I can’t see that stated in post 1.

 

[Thermofox]

Be careful not to double count the work done against friction or by the spring.
What are you asked to find? I can’t see that stated in post 1.

2)The velocities of both the block and the slab, when the block reaches the left edge of the slab.

 

[haruspex]

2)The velocities of both the block and the slab, when the block reaches the left edge of the slab.

Hmm.. I should have looked harder.

In post #9 you tried to apply work conservation separately to two parts of the system. That is risky because the parts interact. Safer to apply it to the whole system.
Friction is to do with relative motion of two surfaces in contact. The work done is related to that relative motion.

What other conservation law can be applied?

 

[Thermofox]

Hmm.. I should have looked harder.

In post #9 you tried to apply work conservation separately to two parts of the system. That is risky because the parts interact. Safer to apply it to the whole system.
Friction is to do with relative motion of two surfaces in contact. The work done is related to that relative motion.

What other conservation law can be applied?

I can only think about momentum, but I don't think that there is a conservation of it. Because there is friction which is an external force.

 

[Steve4Physics]

I can only think about momentum, but I don't think that there is a conservation of it. Because there is friction which is an external force.

Consider the slab, block and spring as a single system. Friction and the forces exerted by the spring are then internal forces within the system.

There is no net external force on the system. What does conservation of momentum tell you about the motion of the system’s centre of mass?

And what does the law of conservation of momentum then tell you about the motion of the system’ internal components?

 

[Thermofox]

There is no net external force on the system. What does conservation of momentum tell you about the motion of the system’s centre of mass?

$m_{cm}v_{cm,i}= m_{cm} v_{cm,f}$?

And what does the law of conservation of momentum then tell you about the motion of the system’ internal components?

$m_b v_{b,i} = m_b v_{b,f}$ and $m_s v_{s,i} = m_s v_{s,f}$?

I'm not that confident in what I wrote.

 

[Thermofox]

Consider the slab, block and spring as a single system. Friction and the forces exerted by the spring are then internal forces within the system.

There is no net external force on the system. What does conservation of momentum tell you about the motion of the system’s centre of mass?

And what does the law of conservation of momentum then tell you about the motion of the system’ internal components?

From what I see, the center of mass has to maintain the same velocity => $v_s + v_b = 0$
Therefore $v_s = -v_b$?

 

[haruspex]

From what I see, the center of mass has to maintain the same velocity => $v_s + v_b = 0$
Therefore $v_s = -v_b$?

No, the masses are different. Stick with momentum conservation.

 

[Thermofox]

No, the masses are different. Stick with momentum conservation.

I'm really struggling here, I can't find a conclusion. Could you give me another hint?

 

[Thermofox]

No, the masses are different. Stick with momentum conservation.

If the momentum of the whole system is conserved then $$(m_b + m_s) v_{initial} = (m_b + m_s) v_{final}$$
Then since at the start the system is still $v_i =0$ and $v_f= v_s + v_b$. That's why I wrote $v_s + v_b = 0$. The mass of the system doesn't change.

 

[Steve4Physics]

I asked "There is no net external force on the system. What does conservation of momentum tell you about the motion of the system’s centre of mass?" You replied:

$m_{cm}v_{cm,i}= m_{cm} v_{cm,f}$?

Yes. For the conceptual understanding, I would express this in words:

$m_{cm}$ is the total mass - which doesn't change. This means $v_{cm,i}=v_{cm,f}$, i.e. the velocity of the centre of mass doesn't change.

We assume the system is stationary before the spring is relased. That means the system's centre of mass remains stationary after the spring is released.

_______________________________

I asked "And what does the law of conservation of momentum then tell you about the motion of the system’ internal components?". You replied:

$m_b v_{b,i} = m_b v_{b,f}$ and $m_s v_{s,i} = m_s v_{s,f}$?

That's not always true.
_______________________________

Here's a different question. A system consists of stationary bomb with no external forces acting on it. The bomb explodes and the system now consists of two unequal parts of masses $m_1$ and $m_2$ with velocities $\vec {v_1}$ and $\vec {v_2}$.

Q1) What are the velocities of the system's centre of mass before and after the explosion?

Q2) How are $m_1, m_2, \vec {v_1}$ and $\vec {v_2}$ related?

 

[Thermofox]

Q1) What are the velocities of the system's centre of mass before and after the explosion?

Before the explosion $v_i=0$, after the explosion $v_f=v_1 + v_2$

Q2) How are $m_1, m_2, \vec {v_1}$ and $\vec {v_2}$ related?

$m_{bomb} \vec {v_i}= m_1 \vec {v_1} + m_2 \vec {v_2}$

 

[Steve4Physics]

Before the explosion $v_i=0$

Yes. Before the explosion the system's centre of mass is stationary.

, after the explosion $v_f=v_1 + v_2$

No. There is no external force on the system. So the system's centre of mass can not accelerate - it will remain stationary. $v_f=0$.

You need to take a step back and understand the basic physics. Remember velocity and momentum are vectors. Try this problem:

A stationary bomb has mass 10kg. No external forces act on it. The bomb explodes into two pieces. One piece has mass 4kg and moves at 12m/s to the right. Use right=positive, left = negative for vector-values. Without any equations or symbols:
a) What is the total momentum after the explosion?
b) What is the momentum of the 4kg piece?
c) What is the momentum of the 6kg piece?
d) What is the velocity of the 6kg piece?

It's my bedtime now so I'm closing down for the night!

 

[haruspex]

If the momentum of the whole system is conserved then $$(m_b + m_s) v_{initial} = (m_b + m_s) v_{final}$$

I assume $v_{initial}$ and $v_{final}$ are velocities of the common mass centre, in which case …

$v_f= v_s + v_b$.

… no. You cannot add velocities of the different bodies like that because the masses are different. What you can add are their momenta.

 

[Thermofox]

Yes. Before the explosion the system's centre of mass is stationary.


No. There is no external force on the system. So the system's centre of mass can not accelerate - it will remain stationary. $v_f=0$.

You need to take a step back and understand the basic physics. Remember velocity and momentum are vectors. Try this problem:

A stationary bomb has mass 10kg. No external forces act on it. The bomb explodes into two pieces. One piece has mass 4kg and moves at 12m/s to the right. Use right=positive, left = negative for vector-values. Without any equations or symbols:
a) What is the total momentum after the explosion?
b) What is the momentum of the 4kg piece?
c) What is the momentum of the 6kg piece?
d) What is the velocity of the 6kg piece?

It's my bedtime now so I'm closing down for the night!

a) 0 kgm/s
b) 48 kgm/s
c)-48 kgm/s
d)-8 m/s

 

[Thermofox]

I assume $v_{initial}$ and $v_{final}$ are velocities of the common mass centre, in which case …

… no. You cannot add velocities of the different bodies like that because the masses are different. What you can add are their momenta.

$m_s v_{s,f} + m_b v_{b,f}=0$ because the system was initially stationary.

 

[Steve4Physics]

a) 0 kgm/s
b) 48 kgm/s
c)-48 kgm/s
d)-8 m/s

That's right. Now think of the slab and block (from the original question) as 2 parts of a 'bomb'.

The momenta of the 2 parts will be equal in magnitude and opposite in direction. So the velocities (and hence displacements) of the 2 parts are related.

 

[Thermofox]

That's right. Now think of the slab and block (from the original question) as 2 parts of a 'bomb'.

The momenta of the 2 parts will be equal in magnitude and opposite in direction. So the velocities (and hence displacements) of the 2 parts are related.

$m_b v_b= - m_s v_s$ => $v_b= \frac {-m_s v_s} {m_b}$ => $x_b= \frac {-m_s x_s} {m_b}$?

Since i know the masses, $v_b= -4v_s$. Now I just need to determine $v_b (\text {or}\space v_s)$ and I can do that by using the energetic balance: $$\frac 1 2 m_s v_s^2 -2 m_b v_s^2 - \frac 1 2 k \Delta l^2 = - W_{\text{friction}}$$

Is the work done by friction, $W_f= \Delta l \mu_d g m_b$?

 

[haruspex]

$m_s v_{s,f} + m_b v_{b,f}=0$ because the system was initially stationary.

Yes.

 

[Thermofox]

Yes.

Now can I say that $W_{\text{friction}}= \Delta l \mu_d g m_b$?
I'm not sure because the work is done on 2 surfaces that are moving.

 

[Thermofox]

That's right. Now think of the slab and block (from the original question) as 2 parts of a 'bomb'.

The momenta of the 2 parts will be equal in magnitude and opposite in direction. So the velocities (and hence displacements) of the 2 parts are related.

This means that the center of mass of the system is not moving and I can visualize that because the block and the slab are moving in opposite directions. But what would happen when the block falls from the slab. Is our system still valid or as soon as they detach, they can no longer be considered as a system?

 

[haruspex]

Now can I say that $W_{\text{friction}}= \Delta l \mu_d g m_b$?

Yes.

I'm not sure because the work is done on 2 surfaces that are moving.

Depending on the frame of reference you choose, you could consider that the block moves $x$ one way and the slab moves $\Delta l-x$ the other way. So the block does work $\mu_kxm_bg$ against friction while the slab does work $\mu_k(\Delta l-x)m_bg$ against friction. The total work done against friction is always $\mu_k\Delta lm_bg$.

But what would happen when the block falls from the slab.

You understand that you do not need to consider that, right?

Is our system still valid or as soon as they detach, they can no longer be considered as a system?

A system is whatever you define it to be.
Certainly it will behave differently after detachment.

 

[Steve4Physics]

But what would happen when the block falls from the slab. Is our system still valid or as soon as they detach, they can no longer be considered as a system?

The question does not ask about what happens when the block falls from the slab. It wants the velocities at the moment before this happens.

 

[Thermofox]

You understand that you do not need to consider that, right?

Yeah, I perfectly do. I was just a thought I had, nothing more.

Depending on the frame of reference you choose, you could consider that the block moves $x$ one way and the slab moves $\Delta l-x$ the other way. So the block does work $\mu_kxm_bg$ against friction while the slab does work $\mu_k(\Delta l-x)m_bg$ against friction. The total work done against friction is always $\mu_k\Delta lm_bg$.

Ok now I understand it, Thank you immensely!

 

[Thermofox]

The question does not ask about what happens when the block falls from the slab. It wants the velocities at the moment before this happens.

Yeah I do realize that, It was just an "if case" I had.