Need help with Incline Plane Problem

  • Thread starter jord12321
  • Start date
  • #1
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Need help with Incline Plane Problem!!

Homework Statement


A 10 kg block of ice slides down a ramp 20 m long, inclined at 10 degrees to the horizontal.
a) If the ramp is frictionless, what is the acceleration of the block of ice? (I was able to do this question the answer is 1.7 m/s²)

b) If the coefficient of kinetic friction is .10, how long will it take the block to slide down the ramp, if it starts from rest?

So right now I know I have...
m= 10 kg
d= 20 m
sin10
cos10
v1= 0 m/s
a= 1.7 m/s²


Homework Equations


I'm really not sure what to do.


The Attempt at a Solution


I'm not sure what to do, I need to know where to start!
 

Answers and Replies

  • #2


Homework Statement


A 10 kg block of ice slides down a ramp 20 m long, inclined at 10 degrees to the horizontal.
a) If the ramp is frictionless, what is the acceleration of the block of ice? (I was able to do this question the answer is 1.7 m/s²)

b) If the coefficient of kinetic friction is .10, how long will it take the block to slide down the ramp, if it starts from rest?

So right now I know I have...
m= 10 kg
d= 20 m
sin10
cos10
v1= 0 m/s
a= 1.7 m/s²


Homework Equations


I'm really not sure what to do.


The Attempt at a Solution


I'm not sure what to do, I need to know where to start!

To start with, you should probably draw a FBD of the block on the ramp. :smile:
 
  • #3
13
0


Ok i did that, I just don't know what to solve for first.
 
  • #4


Alright, so you drew your FBD.

You have a Weight force going down.
A normal force going perpendicular to the ramp's surface.
A friction force going the opposite direction of motion.

Correct?

Now..... 2 questions for you.

[itex]\Sigma[/itex]F[itex]_{y}[/itex]= ?
[itex]\Sigma[/itex]F[itex]_{x}[/itex]= ?
 
  • #5
13
0


correct,
fx = sin10mg = sin10 x 10 x 9.8 = 17
fy = cos10mg = cos10 x 10 x 9.8 = 96
 
  • #6


Where is your friction force? I can't tell if you have one or not.
 
  • #7
13
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What do you mean? Didn't i just solve for parallel and perpendicular force?
 
  • #8


I might be missing something myself; running on very little sleep here. Sorry.
 
  • #9
13
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I don't have the force of friction yet.
 
  • #10
13
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I was able to get friction because fn = f perpendicular. So i did .10 x 96 and got friction to equal -9.6
 
  • #11


Ok, so after looking at this myself....

I get.

[itex]\Sigma[/itex]F[itex]_{x}[/itex]=mg*sin[itex]\Theta[/itex] - [itex]\mu[/itex]mg*cos[itex]\Theta[/itex]

I'm still working on Fy.
 
  • #12
13
0


Ok, thanks for your help!
 
  • #13


I was able to get friction because fn = f perpendicular. So i did .10 x 96 and got friction to equal -9.6

Friction = [itex]\mu[/itex]mg*cos[itex]\Theta[/itex]

The reasoning:

N = -mg*cos[itex]\Theta[/itex]

So, friction = -.10*98*cos(10) = -9.65 N

Correct?


You're welcome, but I don't know how much help I was. :/ Fy always gets me in these problems.
 
  • #14
13
0


Friction = [itex]\mu[/itex]mg*cos[itex]\Theta[/itex]

The reasoning:

N = -mg*cos[itex]\Theta[/itex]

So, friction = -.10*98*cos(10) = -9.65 N

Correct?


You're welcome, but I don't know how much help I was. :/ Fy always gets me in these problems.

I got the same answer as you.
 
  • #15


I got the same answer as you.
Right. And you can find acceleration from there, correct?
 
  • #16
13
0


Correct, but I'm suppose to find time.
 
  • #17


Then you use the acceleration and plug it into a kinematic equation.

You have delta x, acceleration, and an initial velocity.

So, you have to find your time using.....

[itex]\Delta[/itex]x = V[itex]_{i}[/itex]t + 1/2(a)(t[itex]^{2}[/itex])

Please correct me if I'm wrong.
 
  • #18
13
0


You are right!!! Thank you so much for helping!!
 
  • #19


You're very welcome!
 

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