Conservative forces on a sliding block

In summary, Homework Equations state that: -E-mec 1 (object moving) = K1 + U1-E-mec 2 (object stopped) = K2 + U2 + energy lost to friction
  • #1
peaceandlove
67
0

Homework Statement


A 4.5 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.3 m/s, the height difference is h = 1.0 m, and μk = 0.613. Find d.


Homework Equations


E-mec 1 (object moving) = K1 + U1
E-mec 2 (object stopped) = K2 + U2 + energy lost to friction


The Attempt at a Solution


K1 + U1 - energy lost to friction = K2 + U2
I said that K1=(1/2)mv^2 and U1=mgy and the energy lost to friction is (fk)d; however, I don't know how to get the right side of the equation. I tried using the same formulas and I came up with 1.6313, but apparently that is incorrect.
 
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  • #2
peaceandlove said:
K1 + U1 - energy lost to friction = K2 + U2
This looks good. So you explicitly we have:

[tex]\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + R\mu_k d[/tex]

And you want to find the distance, i.e. you want to solve this equation for d. What's the first thing you usually do when solving an equation?
 
  • #3
The first thing I would do is plug in all the values I know.

m=4.5kg
v1=5.3m/s
g=9.8m/s^2
h1=1m (not entirely sure about this one...)
v2=(I would think it to still be 5.3m/s, but I could be wrong)
h2=0 (I think...)
R=(no clue.)
Uk=0.613
 
  • #4
peaceandlove said:
m=4.5kg
v1=5.3m/s
g=9.8m/s^2
Good.
peaceandlove said:
h1=1m (not entirely sure about this one...)
h2=0 (I think...)
Don't worry about these, we'll get to these later.
peaceandlove said:
v2=(I would think it to still be 5.3m/s, but I could be wrong)
Hint: You are looking for the distance d when the object has stopped.
peaceandlove said:
R=(no clue.)
What is the expression for the maximum frictional force on an object?
peaceandlove said:
Uk=0.613
Good.

Personally, I wouldn't plug in any of the values in just yet. Since we're solving for d I would try to make d the subject of the expression.
 
  • #5
So would v2 be 0 and R=(Us)(Fn)?
 
  • #6
And d=((1/2)m(v1)^2+mg(h1)-(1/2)m(v2)^2-mg(h2))/R(Uk).
 
  • #7
peaceandlove said:
So would v2 be 0 and R=(Us)(Fn)?
Sounds good to me :approve:
peaceandlove said:
And d=((1/2)m(v1)^2+mg(h1)-(1/2)m(v2)^2-mg(h2))/R(Uk).
Yup. However, you can write it in a somewhat nicer fashion:

[tex]\begin{align*}
d & = \frac{1}{R\mu_k}\left\{\frac{1}{2}mv_1^2 + mgh_1 - \frac{1}{2}mv_2^2 -mgh_2\right\} \\
& = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\left(h_1-h_2\right)\right\} \\
& = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\Delta h\right\}\end{align*}[/tex]

Does that help?
 
  • #8
One last thing, what values do we plug in for (Us) and (Fn)? I got (Fn) to be 44.1 (although I could be wrong), but I don't know how to find (Us).
 
  • #9
peaceandlove said:
One last thing, what values do we plug in for (Us) and (Fn)? I got (Fn) to be 44.1 (although I could be wrong), but I don't know how to find (Us).
Sorry, I misread your previous post. R is simply Fn= 44.1 N.

As a side note, take care with the sign of [itex]\Delta h[/itex], recall it's definition [itex]\Delta h = h_1-h_2[/itex].
 
  • #10
Hootenanny said:
Good.

Don't worry about these, we'll get to these later.

Hint: You are looking for the distance d when the object has stopped.

What is the expression for the maximum frictional force on an object?

Good.

Personally, I wouldn't plug in any of the values in just yet. Since we're solving for d I would try to make d the subject of the expression.

Oh, and you never got back to clarifying what h1 and h2 are.
 
  • #11
peaceandlove said:
Oh, and you never got back to clarifying what h1 and h2 are.
What do h1 and h2 represent?
 
  • #12
The change in y, right? So h1 is 1 and h2 is 0?
 
  • #13
peaceandlove said:
The change in y, right? So h1 is 1 and h2 is 0?
Almost. h1 and h2 represent the initial and final heights respectively. Hence, [itex]\Delta h = h_1-h_2[/itex] represent the change in height. However, you should note that the question states that the final height is greater than the initial height.
 
  • #14
Oh, so it's the other way around? Meaning (delta)h=-1?
 
  • #15
peaceandlove said:
Oh, so it's the other way around? Meaning (delta)h=-1?
Yes. :approve:
 
  • #16
Thank you so much for your help!
 
  • #17
peaceandlove said:
Thank you so much for your help!
A pleasure :smile:
 

Related to Conservative forces on a sliding block

1. What are conservative forces?

Conservative forces are those that do not dissipate energy as they act on an object. This means that the work done by these forces is independent of the path taken by the object and only depends on the initial and final positions of the object.

2. What are some examples of conservative forces?

Examples of conservative forces include gravity, electric and magnetic forces, and elastic forces.

3. How do conservative forces affect a sliding block?

Conservative forces on a sliding block will only affect the block's motion if they are unbalanced. If the forces are balanced, the block will continue to slide at a constant speed or remain stationary if at rest.

4. How can we determine if a force is conservative on a sliding block?

A force is conservative if it satisfies the condition of path independence. This means that the work done by the force is the same regardless of the path taken by the block. Mathematically, this can be expressed as the integral of the force over the path being equal to the difference in potential energy between the initial and final positions of the block.

5. What is the relationship between conservative forces and energy?

Conservative forces are closely linked to the concept of potential energy. As the work done by these forces is path independent, any change in the block's potential energy is only dependent on its initial and final positions. This means that the work done by conservative forces can be stored as potential energy, which can then be converted back into kinetic energy as the block moves along a path.

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