Block sliding down a ramp problem

  • #1
~christina~
Gold Member
715
0

Homework Statement



The path that a block slides along is frictionless until the block reaches the section L shown in the figure below. The length L = 0.75 m begins at height h = 2.0 m on a ramp of angle θ = 30º. In that section, the coefficient of kinetic friction is 0.40. The block passes through point A with a speed of 8.0 m / s.

If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

http://img49.imageshack.us/img49/633/frictionboxpu8.jpg [Broken]

Homework Equations



[tex]K_A = U_g + K_f -f_kd [/tex]

[tex]f_k=\mu_kmg [/tex]


The Attempt at a Solution



L (friction spot)= 0.75m
[tex]h_f= 2.00m [/tex]

[tex]\theta= 30^o [/tex]

[tex]\mu = 0.40 [/tex]

[tex] V_A= 8.0m/s [/tex]


I was thinking conservation of momentum...

do I take A. to be the initial V? I was thinking it would be but not sure. (continuing as if I'm correct)

[tex]K_A + U_g= U_g + K_f -f_kd [/tex]

[tex]0.5 mv_A^2 + 0 = mgd sin 30^o + 0 - \mu mgL [/tex]

Is this correct?
or does the distance for the friction have to include the angle?
and thus be

[tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mgLsin 30^o [/tex]
 
Last edited by a moderator:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Find the velocity of the block at h = 2m. Then use the conservation of energy to find the final velocity at B.
While going up the block encounters retardation due gravity and frictional force. While calculating the final velocity take these in account. If you get negative answer then the block won't reach the point B.
 
Last edited:
  • #3
~christina~
Gold Member
715
0
was my equation correct though ? (to find the height)

I was really curious to know if I needed to make the frictional force include the angle I was given.

[tex]0.5 mv_A^2 + 0 = mgd sin 30^o + 0 - \mu mgLsin 30^o [/tex]
 
Last edited:
  • #4
rl.bhat
Homework Helper
4,433
7
Frictional force = mu*mg*cos(theta). So the total work done along the ramp = [mu*mg*cos(theta) + mg*sin(theta)]*L
 
  • #5
~christina~
Gold Member
715
0
Frictional force = mu*mg*cos(theta). So the total work done along the ramp = [mu*mg*cos(theta) + mg*sin(theta)]*L

why is it all multiplied by L ?

L= only the part of ramp with friction.

d= total distance traveled by block
 
  • #6
rl.bhat
Homework Helper
4,433
7
Because the frictional force acts only along L, not through out the displacement. Since we don't know the total distance d try to work out with h and L. In your equation work done against the friction should be mu*mg*cos(theta)*L.
 
  • #7
~christina~
Gold Member
715
0
[tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mg L cos 30^o [/tex]

is it fine now?

well assuming it is since nobody answered me.

WAIT A SEC I don't have the mass of the block...Does it cancel out??

Mass matters so how would I find it if the mass isn't given??
 
Last edited:
  • #8
rl.bhat
Homework Helper
4,433
7
WAIT A SEC I don't have the mass of the block...Does it cancel out??
Yes.
Now what is y?
 
  • #9
~christina~
Gold Member
715
0
y= 2m

I figured that for the first part where I go and calculate the final v at the height 2m

How would I calculate the v at the final point B ? I'm not quite sure how the eqzn would look.


[tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

I think this is fine to calculate the v at the 2m point on the ramp but I'm not sure.

I got v= 4.98m/s

but how would I find the v at B (to see whether it's negative or not?)


Is it

[tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mg L cos 30^o [/tex]


I have to get to sleep...I'll check in the morning
 
Last edited:
  • #10
rl.bhat
Homework Helper
4,433
7
Height of B with respect to A ...H = h + l = 2 + 0.75sin(30).
Now you can wright 0.5mVA^2 - 0.5mVB^2 = mgH + mu*mg*cos(30)*L.
Solve for VB.
 
Last edited:
  • #11
~christina~
Gold Member
715
0
so when I wrote this => [tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

I guess it was okay right? (I was thinking I needed a sin or cos somewhere)

for the

Height of B with respect to A ...H = h + l = 2 + 0.75sin(30).
Now you can wright 0.5VA^2 - 0.5VB^2 = mgH + mu*mg*cos(30)*L.
Solve for VB.
I don't have m so how would that work? I don't think it will cancel out this time.

Help
 
  • #12
rl.bhat
Homework Helper
4,433
7
I have edited my mail which you have quoted.
The correct expression is: 0.5mVA^2 - 0.5mVB^2 = mgH + mu*mg*cos(30)*L.
When you cancell m, you get...0.5*8^2 - 0.5*VB^2 = 9.8*(2 + 0.375) + 0.4*9.8*0.866*0.75. Solve it and find the value of VB.
 
  • #13
~christina~
Gold Member
715
0
I have edited my mail which you have quoted.
The correct expression is: 0.5mVA^2 - 0.5mVB^2 = mgH + mu*mg*cos(30)*L.
When you cancell m, you get...0.5*8^2 - 0.5*VB^2 = 9.8*(2 + 0.375) + 0.4*9.8*0.866*0.75. Solve it and find the value of VB.

for the initial velocity at [tex]V_{2.00m}[/tex] I got 4.98m/s with the equation

[tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

why did I get that if I don't use that as a initial velocity for V at h = 2.00 but instead use the VA which is bellow ramp to find V at the point B?
 
Last edited:
  • #14
rl.bhat
Homework Helper
4,433
7
Initially I suggested that keeping in mind to use the formula:
0.5*V2m^^2 - 0.5*VB^2 = 9.8* 0.375 + 0.4*9.8*0.866*0.75. After seeing your equation I thought it is better to do at one stretch. In your equation the error is you have assumed VB = 0 and you have taken work done against friction negative.
 
  • #15
~christina~
Gold Member
715
0
Okay Thank You :smile:
 

Related Threads on Block sliding down a ramp problem

Replies
8
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
18
Views
858
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
12K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
0
Views
4K
Replies
3
Views
5K
Replies
4
Views
8K
Replies
6
Views
5K
Top