Having trouble solving Nonconservative Force problem

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In summary: I incorrectly read it the first time through. ## In summary, the problem involves a 0.75-kg block sliding on a frictionless surface with an initial speed of 2.0 m/s. The block then slides over a rough area and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17. The goal is to find the speed of the block after it passes over the rough surface. The solution involves setting up an equation using energy methods and taking into account the work done by the friction force over the distance traveled. The book's answer of 0.82 m/s is correct, as confirmed by solving the equation symbolically.
  • #1
Iconic
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Homework Statement



A 0.75-kg block slides on a frictionless surface with a speed of 20 m/s. It then slides over a rough area 1.0m in length and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17.What is the speed of the block after it passes across the rough surface?

m = 0.75 kg
x(during friction being applied)= 1.0m
Coefficient of Friction = 0.17
Initial Velocity = 20 m/s
Gravity = 9.8​



Homework Equations



-Equations I used -
() - means multiplying.

Normal Force = Mass(Gravity)
-Just used the Fx component since Fy equals 0-
=
Friction Force= Coefficient(Normal Force)(Distanceof the friction)

K = 1/2mv2

Ei = Ef
=
Ki - Friction Force = Kf
=
1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2



The Attempt at a Solution


I first realized what I needed to find, which was the Final Velocity.

Then set up my equation.
1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

Then plugged in the numbers.
1/2(.75kg)(20m/s)-(.17f)(.75kg)(9.8m/s2)(1.0m) = 1/2(.75kg)(v2)

Simplified.
6.2505 = 3.75 v2

Simplified again.
1.6668 = v2

Squared.
1.291 = v

Books Answer
However, When I looked at the book they got the answer 0.82m/s and got it by doing...
v=√(InitialV2) - 2 (Coefficient of friction)(Gravity)(Distance))
* Note - all of this is under the square root *
v = √(2.0)2-2(0.17)(9.8)(1.0)
=0.82 m/s

My question is... What did I do wrong and what did the book do correctly? I would really love a good explanation on this as It's important for me to grasp this concept. The answer is there but I want to know HOW and WHY. Thank you in advance whoever takes the time to answer this it is greatly appreciated and your positivity will be rewarded.

Kind regards, A lost soul.
 
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  • #2
You used energy methods nicely, but made a math error by not squaring the 20...check your work again and you should see it should have been K_i = 1/2 m (20)^2. The book used the kinematic equation after first solving for the acceleration during the friction phase. Either way is ok.
 
  • #3
Hi Iconic, Welcome to Physics Forums.

Iconic said:

Homework Statement



A 0.75-kg block slides on a frictionless surface with a speed of 20 m/s. It then slides over a rough area 1.0m in length and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17.What is the speed of the block after it passes across the rough surface?

m = 0.75 kg
x(during friction being applied)= 1.0m
Coefficient of Friction = 0.17
Initial Velocity = 20 m/s
Gravity = 9.8​



Homework Equations



-Equations I used -
() - means multiplying.

Normal Force = Mass(Gravity)
-Just used the Fx component since Fy equals 0-
The usual convention is to place the x-axis horizontal and the y-axis vertical. Thus the normal force points along the y-axis, normal (perpendicular) to the Earth's surface (which is why it's called the "normal force").

Friction Force= Coefficient(Normal Force)(Distanceof the friction)
Nooo. Force is not work. When you multiply a force by a distance you get work done by that force moving something through that distance. The friction coefficient multiplied by the normal force yields the friction force. ##\mu_k F_n = F_f##.

K = 1/2mv2

Ei = Ef
=
Ki - Friction Force = Kf
Again, force is not energy. You want an energy term there. In particular you want the work done by the Friction Force on the mass m over the 1.0 meter distance.

=
1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2
That last line is good, except you should use different variables for the initial and final velocities.

The Attempt at a Solution


I first realized what I needed to find, which was the Final Velocity.

Then set up my equation.
1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

Then plugged in the numbers.
1/2(.75kg)(20m/s)-(.17f)(.75kg)(9.8m/s2)(1.0m) = 1/2(.75kg)(v2)

Simplified.
6.2505 = 3.75 v2

Simplified again.
1.6668 = v2

Squared.
1.291 = v

Books Answer
However, When I looked at the book they got the answer 0.82m/s and got it by doing...
v=√(InitialV2) - 2 (Coefficient of friction)(Gravity)(Distance))
* Note - all of this is under the square root *
v = √(2.0)2-2(0.17)(9.8)(1.0)
=0.82 m/s
That value is highly suspicious. The formula is fine but the quoted result is obviously wrong. Maybe a typo in the book? Consider that the block starts out going 20 m/s and passes over a mere 1.0 meter of barely rough surface (0.17 as a friction coefficient is pretty small). I'd expect a value much closer to the initial speed.

EDIT: Re-reading the numbers through your post I see that you meant the initial speed to be 2.0 m/s, not 20 m/s! Big difference! The book's answer is correct. It' really is easier to follow symbols than numbers with decimals when parsing another's work.

My question is... What did I do wrong and what did the book do correctly? I would really love a good explanation on this as It's important for me to grasp this concept. The answer is there but I want to know HOW and WHY. Thank you in advance whoever takes the time to answer this it is greatly appreciated and your positivity will be rewarded.

Kind regards, A lost soul.

It pays to leave the problem in symbolic form as long as possible. During your workings you wrote:

1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

Let me translate that into slightly more compact form, writing out the normal force as it was formulated:
$$\frac{1}{2} m v_i^2 - (\mu_k m g) d = \frac{1}{2} m v_f^2$$
What do you get if you multiply though by 2 and divide through by m? Look familiar?
 
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  • #4
I see what you are saying however, after realizing my mistake I now tried to solve the equation with this in mind and still ended up with an incorrect result.

1/2(.75kg)(20m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = 1/2(.75kg)(v2)

Simplified.
150-1.2495 = .375(v2)
Simplified Again.
148.7505 = .375(v2)
Step 3
396.668 = v2
Step 4
√(396.668) = v
Step 5
19.91 = v

Still not right =/
 
  • #5
Iconic: Check the edit that I'd added to my previous post.

19.92 m/s (rounded) would be correct if the initial speed was 20 m/s :smile:
 
  • #6
Sorry, for replying so late I'm just taking my time with these posts to ensure I didn't mess up anywhere. Just read your post gneill, THANK YOU very much for that broad and well written explanation. Couple things I need to say though, there might have been a typo, however; after double checking it still says 20 m/s in the question and 2.0 when the book solves for Vf:bugeye:. But anyways, when you multiply 2 from 1/2mvf2 on to the other side and divide out m is this how the equation should look?

1/2mvi2-Coefficient(Normal Force)(Distance)= 1/2mv2
*2
mvi2-Coefficient(Normal Force)(Distance)= mv2
/m
vi2-Coefficient(Normal Force)(Distance)= v2

Solving: Assuming Vi was 20

(20m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = v2

400 - 1.2495 = v2

398.7505 = v2

√(398.7505)

v=19.96

Solving: Assuming Vi was 2.0

(2.0m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = v2

4 - 1.2495 = v2

2.7505 = v2

√(2.7505)

v= 1.658

Assuming this was a typo... I'm not getting the right answer =/ Was my algebra incorrect? I feel like I'm not manipulating the equation correctly.:frown:

EDIT - wanted to point out that I realize this equation is supposed to look similar if not exactly like the kinematic equation
vf2 = vi2 + 2a Delta X
But I'm horrid with algebra and am having a difficult time getting it there.
 
Last edited:
  • #7
Iconic said:
Sorry, for replying so late I'm just taking my time with these posts to ensure I didn't mess up anywhere. Just read your post gneill, THANK YOU very much for that broad and well written explanation. Couple things I need to say though, there might have been a typo, however; after double checking it still says 20 m/s in the question and 2.0 when the book solves for Vf:bugeye:. But anyways, when you multiply 2 from 1/2mvf2 on to the other side and divide out m is this how the equation should look?

1/2mvi2-Coefficient(Normal Force)(Distance)= 1/2mv2
*2
mvi2-Coefficient(Normal Force)(Distance)= mv2
Ooops! You forgot the middle term. When you multiply through or divide through an entire equation you must include every term.

[SNIP --- the above affected the excised lines]
:
:
:
v= 1.658

Assuming this was a typo... I'm not getting the right answer =/ Was my algebra incorrect? I feel like I'm not manipulating the equation correctly.:frown:
Yup. You missed multiplying the friction work term by two. Give it another go.

EDIT - wanted to point out that I realize this equation is supposed to look similar if not exactly like the kinematic equation
vf2 = vi2 + 2a Delta X
But I'm horrid with algebra and am having a difficult time getting it there.
Practice will polish your skills. Everyone climbs the same mountain :smile:
 
  • #8
So what I really do is...

1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

Multiply the two from all the variables on the opposing side? Not just 1

mv2- 2Coefficient(Normal Force)(Distance)= mv2

Then Mass is divided canceling out every where

v2- 2Coefficient(Normal Force)(Distance)= v2

This makes sense now. Only 1 question and I know this is a dumb one but I might as well ask:redface:. How come Normal Force and Distance isn't multiplied by 2 as well? Is that because that's supposed to equal Work of Friction so it's essentially 1 variable?
 
  • #9
You are still having algebra problems. You forgot to divide the middle term by m when you divided the other terms by m. The normal force is mg. the friction force is (u)mg . The magnitude of the work done by friction is (u)mgd. Be sure to divide that term by m also, and the equation will be exactly the same as the kinematic equation, as it should be.
 
  • #10
PhanthomJay said:
You are still having algebra problems. You forgot to divide the middle term by m when you divided the other terms by m. The normal force is mg. the friction force is (u)mg . The magnitude of the work done by friction is (u)mgd. Be sure to divide that term by m also, and the equation will be exactly the same as the kinematic equation, as it should be.


Ahh, thank you for clearing that up I was having difficulty with defining some forces but that explains it. So it's not the friction force we were trying to get but the magnitude of that frictional force over 1.0m. Thank you so much that really helped!:approve:

v2- 2(u)gd= v2
 
  • #11
Iconic

You said
v2- 2(u)gd= v2
when you meant to say
vi2- 2(u)gd= vf2
. You are given vi = 2.0 m/s, and you are trying to solve for vf.

The proper way to do the algebra is to isolate vf to get the equation
[itex]v_f = \sqrt{v_i^2 -2ugd}[/itex], from which
[itex]v_f = \sqrt{4 - 2(.17)(9.8)(1.0)}[/itex] or thus
[itex]v_f = \sqrt{4 - 3.332}[/itex]
[itex]v_f = \sqrt {.668} = 0.82 m/s[/itex]

Now if you find the algebra difficult, you can instead 'plug and chug':

[itex]v_i^2 -2ugd = v_f^2[/itex]
[itex]4 - 3.332 = v_f^2[/itex]
[itex]0.668 =v_f^2[/itex]
[itex]v_f = 0.82 m/s[/itex]

This latter method is subject to errors of its own, and loses some information, but sometimes it makes the algebraic manipulations easier, in a pinch.
 
  • #12
Oops, I forgot to add the subscript i and f on that last equation but what you said was exactly what I meant. Again thanks, I think the only real issue I had with this problem was figuring out how we went from

1/2mvi2-2(u)mgd=1/2mvf2

to

vi2- 2(u)gd= vf2

But I understand now that you're supposed to manipulate the equation for the variable you want then solve. I didn't know that you had to multiply the 2 and cancel out the M in not just one but all variables- that's what the issue was and I get it now.:smile:
 
  • #13
Iconic said:
Oops, I forgot to add the subscript i and f on that last equation but what you said was exactly what I meant. Again thanks, I think the only real issue I had with this problem was figuring out how we went from

1/2mvi2-2(u)mgd=1/2mvf2

to

vi2- 2(u)gd= vf2

But I understand now that you're supposed to manipulate the equation for the variable you want then solve. I didn't know that you had to multiply the 2 and cancel out the M in not just one but all variables- that's what the issue was and I get it now.:smile:
I think you understand the algebra now, but note again your typo: You wrote
1/2mvi2-2(u)mgd=1/2mvf2

to

vi2- 2(u)gd= vf2
that red 2 was not in the original equation , you meant to write
1/2mvi2-(u)mgd=1/2mvf2

to

vi2- 2(u)gd= vf2
by multiplying all terms by 2 and then dividing all terms by m. Just a typo. Good work.
 

1. What are nonconservative forces?

Nonconservative forces are external forces that cannot be described by a potential energy function. They are generally forces that involve friction, air resistance, or other dissipative forces that result in a loss of mechanical energy.

2. Why are nonconservative forces important to consider?

Nonconservative forces play a crucial role in determining the total work done on an object and, therefore, affect its overall energy. They can also cause changes in an object's trajectory and lead to different outcomes in a problem.

3. How do I identify nonconservative forces in a problem?

Nonconservative forces can be identified by considering whether the work done by the force depends on the path taken, rather than just the initial and final positions. If so, the force is likely nonconservative.

4. What are some strategies for solving problems involving nonconservative forces?

One strategy is to consider the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. Another approach is to use the concept of potential energy, which can be used to calculate the work done by conservative forces and then subtract from the total work done to determine the work done by nonconservative forces.

5. How can I check if my solution to a nonconservative force problem is correct?

You can check your solution by ensuring that energy is conserved (i.e. the total energy before and after the problem is the same) and by checking that the final position and velocity of the object match the expected values. Additionally, you can try solving the problem using different methods to see if you get the same result.

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