How Do You Calculate the Final Speed of a Block on an Inclined Plane?

AI Thread Summary
To calculate the final speed of a block on an inclined plane, it's essential to consider both kinetic and potential energy, along with the work done against friction. A free body diagram helps identify the forces acting on the block, and a suitable coordinate system should be established. Newton's second law can be applied, with the equation W_nc = ΔK + ΔU guiding the calculations. The acceleration can be determined using a = g*sin(θ), factoring in friction to find the net force and adjusted acceleration. Ultimately, energy conservation principles can also be utilized to relate potential energy to kinetic energy and work done against friction for accurate speed calculations.
metrocard
Messages
2
Reaction score
0

Homework Statement


http://img687.imageshack.us/img687/8417/screenhunter002.jpg


Homework Equations


I don't know.


The Attempt at a Solution


I don't even know where to begin...
 
Last edited by a moderator:
Physics news on Phys.org
This involves energy right? So kinetic and potential energies have to be considered.
(I just finished learning this, so I might be wrong) :)
Looks like a webwork problem to me... :D
 
A good starting point is to draw a free body diagram and identify the forces acting.

Then a second good step is to introduce a suitable coordinate system One wit hte x-axis parallel to the inclined plane would do. Then choose a suitable y axis.

After you do that, you might want to write down Neton's second law of motion for those coordinate directions.
 
Start with:

Wnc = \DeltaK + \DeltaU

for part 1, you are looking for Vf

this gives you...
-(0.21)(37.8)(7.96) = (1/2)(4.5)(Vf2) - (4.5)(9.8)(4.1)

turns out that Vf is 7.23m/s .. tell me if it's right.

for part 2, you are looking for s, the distance

you use the same: Wnc = \DeltaK + \DeltaU
- (0.21)(4.41)(s) = 0 - (1/2)(4.5)(7.232)
s = 126.99m
 
AEM said:
A good starting point is to draw a free body diagram and identify the forces acting.

Then a second good step is to introduce a suitable coordinate system One wit hte x-axis parallel to the inclined plane would do. Then choose a suitable y axis.

After you do that, you might want to write down Neton's second law of motion for those coordinate directions.
http://img693.imageshack.us/img693/9601/screenhunter002d.jpg
What do I do now?
I found the Force of Friction by doing Ff=uFn. 21*37.8=7.938
How do I find the speed?
Do I use:
(Vf)^2=(Vi)^2+2ad

But then to find the Vf, I need the distance and acceleration.
How do I find the distance and acceleration?

For the acceleration, I did.
a=g*sin(31) to get 5.047. That's the acceleration if there was no friction, right?
Then I used multiplied that by 4.5 to get the net force. I then subtracted the force of friction from the net force to get: 14.78
I divided the new net force by 4.5 to get 3.28 as the new acceleration.
Is that correct?
 
Last edited by a moderator:
metrocard said:
http://img693.imageshack.us/img693/9601/screenhunter002d.jpg
What do I do now?
I found the Force of Friction by doing Ff=uFn. 21*37.8=7.938
How do I find the speed?
Do I use:
(Vf)^2=(Vi)^2+2ad

But then to find the Vf, I need the distance and acceleration.
How do I find the distance and acceleration?

For the acceleration, I did.
a=g*sin(31) to get 5.047. That's the acceleration if there was no friction, right?
Then I used multiplied that by 4.5 to get the net force. I then subtracted the force of friction from the net force to get: 14.78
I divided the new net force by 4.5 to get 3.28 as the new acceleration.
Is that correct?

That is certainly a good start. You can use

V^2_f = V^2_i + 2ad

since the acceleration is a constant.

Another way to do it is to note that the Potential energy at the start goes into kinetic energy at the bottom minus the work to overcome friction. The work to overcome friction is the distance traveled times the frictional force. You can solve this energy equation for the velocity at the bottom and easily check your answer from your other method.

You can also use an energy equation to find out how far it travels. The kinetic energy at the bottom will go into work to overcome friction until the block stops.
 
Last edited by a moderator:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top