Calculating Velocity and Work on an Inclined Plane with and without Friction

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In summary, the block slides down an inclined plane with no friction, but with friction present the final velocity is only 7.0 m/x. Work is done against friction, and the equation for the coefficient of friction in terms of mass, angle, and work is given in the header.
  • #1
saturdaykate
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A 2kg block at the top of an inclined plane at a height h=6m above the ground moves down the plane.

1) Find the value of the velocity when the block is at the bottom of the plane assuming no friction.

2) With friction present, the final velocity is only 7.0 m/x at the bottom of the inclined plane. Find work done against friction.

3) Find an algebraic expression for the coefficient of friction in therms of mass (m), the angle (theta), and the work (W), for the block sliding to the bottom of the inclined plane
 
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  • #2
you can't just post your homework... show you've tried to solve it.
 
  • #3
I have been trying at this stupid problem all day. It was on a test I took, but my whole class couldn't figure out any of the extended response so he let us take it home. For this problem, I kept messing aroudn with the potential energy (mgh) and kinetic energy ((1/2)mv^2) formula but I keep getting really weird answers. I also used the W=energy final - energy initial formulas but that seemed like much of a stretch, and I'm so close to giving up that my eyes are burning but I think that is the point of physics. Any hint or explanation, not an answer, would be appreciated.
 
  • #4
post your equations, let's see why you get "wierd answers".

i think anything beyond use mgh, [tex]\frac{mv^2 }{2}[/tex] and [tex]W=U_{initial}-U_{final}[/tex] will be just solving the problem for you...
 
  • #5
Well, for part 1, I thought that I would use
(MGHf+(1/2)MVf^2)-(MGHi+(1/20MFi^2)
Then, assuming the block starts at rest, you would get
((2kg)(9.8m/s2)(0m)+(1/2)(2kg)(Vf^2))-((2kg)(9.8m/s2)(6m)+(1/2)(2kg)(0m/s))
But what do I set that equal to? I am just completely lost. My teacher is under review and about to lose his tenure but I still can't bear failing his class. I just need a few pointers. Anything.
 
  • #6
well, [tex]U_{final}-U_{initial}[/tex] equals the energy gained by the system. (and - sign means it lost energy)

did the system in your question gained\lost any energy?
 
  • #7
Is the work equal to the kinetic energy added to the potential energy?
 
  • #8
oh wait, work is equal to force times displacement, right? so would i just set that whole mess equal to (2kg)(9.8m/s2)(6m) --- (the potential energy)?
 
  • #9
All the information that I have on the problem was given to you in the header. If you are asking me, I would guess that it lost energy to friction?
 
  • #10
no, kinetic energy + potential energy is not work.
its the total energy of the system (i called it U).
work is the energy difference between the initial and final states.
 
  • #11
so, is using the U equation ( I think my teacher called it the "energy final-energy initial" formula?) going to find me the initial velocity? I AM SO LOST.
 
  • #12
1) Find the value of the velocity when the block is at the bottom of the plane assuming no friction.

2) With friction present, the final velocity is only 7.0 m/x at the bottom of the inclined plane. Find work done against friction.

in the first case you did not have friction - did you lose any energy?

you solved something much like the second question on a different thread you started...
 
  • #13
Okay, so I think I understand question 2. Question one: I did not lose any energy. So... that would mean that my energy at the beginning of the slide woudl equal that at the end of the slide? So would I set potential energy and kinetic energy equal to each other?
 
  • #14
actually
I just took the square root of 2*g*h (rad2(9.8)(6)) and got 10.8444m/s. This is because, like you were trying to tell me I think, that it is not changing in energy? It is conservative, right?
 
  • #15
you got it right
 
  • #16
Thank God! And thank you for your help! I won't even ask you for help on the third part, I think I'll just leave it blank, but thanks for your help...
 

Related to Calculating Velocity and Work on an Inclined Plane with and without Friction

1. How do you calculate the velocity of an object on an inclined plane without friction?

To calculate the velocity of an object on an inclined plane without friction, you can use the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the inclined plane.

2. What is the equation for calculating the work done on an object on an inclined plane with friction?

The equation for calculating the work done on an object on an inclined plane with friction is W = Fd cosθ, where W is the work, F is the force applied, d is the distance the object moves along the inclined plane, and θ is the angle between the force and the displacement.

3. How does friction affect the velocity of an object on an inclined plane?

Friction will decrease the velocity of an object on an inclined plane. This is because friction acts in the opposite direction of the motion of the object, causing it to slow down.

4. Can you have negative work on an object on an inclined plane?

Yes, it is possible to have negative work on an object on an inclined plane. This occurs when the force applied is in the opposite direction of the displacement, resulting in a negative value for work.

5. How does the angle of the inclined plane affect the work done on an object?

The angle of the inclined plane affects the work done on an object by changing the value of cosθ in the work equation. As the angle increases, the value of cosθ decreases, resulting in less work being done on the object.

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