Block Jump from Ramp: Find Speed v

  • Thread starter Thread starter racecar12
  • Start date Start date
  • Tags Tags
    Block Jump Ramp
AI Thread Summary
The discussion revolves around calculating the speed of a block as it leaves a frictionless track and encounters a rough area. The block starts from a height of 4.1m and experiences a change in height of 2.2m before leaving the track horizontally. The conservation of energy principle is applied, where the potential energy at the starting height is converted into kinetic energy minus the work done against friction. The initial calculations led to confusion regarding the inclusion of the distance traveled on the rough surface, but it was clarified that it should not be multiplied in the energy equation. Ultimately, the correct speed of the block when it leaves the track is determined to be approximately 5.96 m/s.
racecar12
Messages
19
Reaction score
1

Homework Statement



A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track
It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 .

What is the speed v of the block when it leaves the track?

Givens:
mass of block =0.401kg
Starting height of block = 4.1m
Horizontal height of block when it encounters the friction = 1.9m
Length of the rough area on the horizontal section of track = 1.3m
Coefficient of friction = 0.3

Homework Equations



Einitial=Efinal
Kinitial +Uinitial = Kfinal +Ufinal

Ffriction=μN
N=W=mg

Δy=h=4.1-1.9=2.2

The Attempt at a Solution



I set Kinitial and Ufinal to zero, yielding:
mgh=0.5mv2d - Ffrictiond
vfinal= [ gh+μNd/0.5(1.3) ]1/2
v=5.96m/s

Just wondering if I went about this correctly,
Thanks
 
Physics news on Phys.org
hi racecar12! :smile:
racecar12 said:
mgh=0.5mv2d - Ffrictiond
vfinal= [ gh+μNd/0.5(1.3) ]1/2

i'm confused :confused:

can you check that, and write it out again, please​
 
Thanks for looking at this post.

Here was my though process as I approached this problem...

We'll just use the conservation of energy theorm for this problem.

Einitial=Efinal
Kinitial+Uinitial=Kfinal+Ufinal

Since the massive object has no initial kinetic energy at the start of the system, call it when the object is at its initial height, it has potential energy equal to mgh. m is the mass of the object, g is the acceleration due to gravity, and h is equal to the distance the object travels. In this case, I have to set h equal to a change in the initial height if the object to its height when it encounters the frictional surface. Δy=h. The left side of the equation now reads: Uinitial=mgh. Or just, mgh

mgh=some final kinetic energy - the work done by friction
We have an equation that relates kinetic energy with velocity...it is 0.5mv2=Kfinal...v2 is also the final velocity, which for this system, is right at the point when object leaves the surface. The object will travel in the air for some unknown distance at some other velocity, but for purposes of this problem, we won't worry about them, because those things are not what this problem is asking for. Now, I'm not sure about multiplying, by length d, but I believe we have to.

I also know that the work done by the frictional force equals μNd. That's because work equals force x distance. We have μ and d given to us. d in this case is the distance the object travels along the rough surface only. N is unknown, but we do know that in this case N =W, which happens to = mg in this case. I may not have told you that the rough surface is flat. The minus sign comes in because the frictional force opposes the motion of the object.

Now our equation looks like this
mgh = 0.5mv2d - 1.532622

Now let's isolate v. My m will cancel when I move everything over to the other side.

(gh + 1.532622)/(0.5)(d) =v2
PlUg it all in and take the square root...v = 5.96 m/s

And sorry that the + is underlined..I'm not sure why it does that when I type it in.
 
racecar12 said:
We have an equation that relates kinetic energy with velocity...it is 0.5mv2=Kfinal...v2 is also the final velocity, which for this system, is right at the point when object leaves the surface. The object will travel in the air for some unknown distance at some other velocity, but for purposes of this problem, we won't worry about them, because those things are not what this problem is asking for. Now, I'm not sure about multiplying, by length d, but I believe we have to.

ah, that's what was puzzling me

no, you don't multiply by d (i've no idea twhere you got that idea from)

KE = 1/2 mv2, that's all!

try again :smile:
 
Yes, you are correct that kinetic energy equals 0.5mv2. However, removing d from the equation takes me much farther from the answer, which I know to be 5.95617m/s. if I proceed with the way that I have thought, I get v to equal 5.96046. Close, but no cigar.

I am making a mistake somewhere in my thought process, but I don't know where it is. Perhaps I'm making multiple mistakes, but I don't know where or what they are. Thanks
 
racecar12 said:
Yes, you are correct that kinetic energy equals 0.5mv2. However, removing d from the equation takes me much farther from the answer, which I know to be 5.95617m/s. if I proceed with the way that I have thought, I get v to equal 5.96046. Close, but no cigar.

I am making a mistake somewhere in my thought process, but I don't know where it is. Perhaps I'm making multiple mistakes, but I don't know where or what they are. Thanks
Can't tell where your error is unless you post the full details of your calculation.
 
haruspex said:
Can't tell where your error is unless you post the full details of your calculation.

Seriously?


racecar12 said:
Thanks for looking at this post.

Here was my though process as I approached this problem...

We'll just use the conservation of energy theorm for this problem.

Einitial=Efinal
Kinitial+Uinitial=Kfinal+Ufinal

Since the massive object has no initial kinetic energy at the start of the system, call it when the object is at its initial height, it has potential energy equal to mgh. m is the mass of the object, g is the acceleration due to gravity, and h is equal to the distance the object travels. In this case, I have to set h equal to a change in the initial height if the object to its height when it encounters the frictional surface. Δy=h. The left side of the equation now reads: Uinitial=mgh. Or just, mgh

mgh=some final kinetic energy - the work done by friction
We have an equation that relates kinetic energy with velocity...it is 0.5mv2=Kfinal...v2 is also the final velocity, which for this system, is right at the point when object leaves the surface. The object will travel in the air for some unknown distance at some other velocity, but for purposes of this problem, we won't worry about them, because those things are not what this problem is asking for. Now, I'm not sure about multiplying, by length d, but I believe we have to.

I also know that the work done by the frictional force equals μNd. That's because work equals force x distance. We have μ and d given to us. d in this case is the distance the object travels along the rough surface only. N is unknown, but we do know that in this case N =W, which happens to = mg in this case. I may not have told you that the rough surface is flat. The minus sign comes in because the frictional force opposes the motion of the object.

Now our equation looks like this
mgh = 0.5mv2d - 1.532622

Now let's isolate v. My m will cancel when I move everything over to the other side.

(gh + 1.532622)/(0.5)(d) =v2
PlUg it all in and take the square root...v = 5.96 m/s

And sorry that the + is underlined..I'm not sure why it does that when I type it in.
 
racecar12 said:
Now our equation looks like this
mgh = 0.5mv2d - 1.532622

You don't need the "d" term and try adding the 1.5 term instead of subtracting since there's thermal energy at the bottom of the ramp.
 
kris2fer said:
You don't need the "d" term and try adding the 1.5 term instead of subtracting since there's thermal energy at the bottom of the ramp.

Hey, thanks for your constructive contribution to my inquiry. I attempted the solution in the way you suggested. I get v to equal 6.3228, which is unfortunately not the correct answer.

I'm wondering if I may have to consider the total length the object travels...like adding Δy to the length of the rough surface.
 
  • #10
racecar12 said:
I get v to equal 6.3228, which is unfortunately not the correct answer.

Really? The equation i get is mgh = 0.5mv2 + umgd
Solve for v and you should get about 5.956173268m/s.
 
  • #11
Oh man, you are correct. I made an algebra mistake while moving everything over to solve for v. Good catch. And thanks a bunch!
 
  • #12
racecar12 said:
Can't tell where your error is unless you post the full details of your calculation.
Seriously?
It had been pointed out to you that mv2d was wrong. You replied that you tried taking out the d and it made matters worse, but you didn't post the calculation. There's also the matter of where 1.532622 came from. I don't see an explanation of that.
 
  • #13
(just got up :zzz:)
racecar12 said:
Seriously?

yes, seriously
racecar12 said:
I made an algebra mistake while moving everything over to solve for v.

that is why we usually insist on you showing your full calculations

please do so next time, and save yourself some time! :smile:
 
Back
Top