How Do You Calculate the Time It Takes for a Block to Stop on a Sloped Ramp?

AI Thread Summary
To calculate the time it takes for a block to stop on a sloped ramp, one must analyze the forces acting on the block, including gravity and friction. The net force can be expressed as F = ma, where the acceleration a is derived from the gravitational component along the ramp minus the frictional force. The equations of motion can then be applied to find the time T it takes for the block to come to rest, using T = -v0/a. The discussion emphasizes the importance of correctly identifying the normal force and its components relative to the ramp's angle. Overall, a clear understanding of the forces and proper application of physics principles leads to the solution.
ChEJosh
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Homework Statement



A block slides down a slope from point O with initial speed V. THe sliding coefficient, mu, brings the block to rest at time T. Using the ramp as the x-axis and the perpendicular as the y-axis, find T.

Homework Equations



F = ma
mu = mg sin(theta)
a=dv/dt

The Attempt at a Solution



I'm taking modern physics, and this is a problem to demonstrate how picking different frames of reference can change a problem. I have to take the axes as the vertical and horizontal after solving this part, butI haven't had general physics in about 3 years, so I'm just not sure how to do the easy part. I think if I get this part I can get the second, I just need a small refresher.

I was thinking somewhere along the lines of this:

ma = mg sin(theta) - mu
dv/dt = g sin(theta) - mu/m
dt = dv/(g sin(theta) - mu/m)

Maybe? But, I'm not sure how I should do that derivative.
Thanks!
 
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ChEJosh said:

Homework Equations



F = ma
That's the one.
mu = mg sin(theta)
mu is the coefficient of friction, so this makes no sense. How does friction relate to normal force?

What you need to do is identify all the forces acting on the block (draw a free body diagram!) and then apply Newton's 2nd law.
 
You're making it too complicated, just find the acceleration and use those old constant acceleration motion equations.
 
Doc Al said:
mu is the coefficient of friction, so this makes no sense. How does friction relate to normal force?

What you need to do is identify all the forces acting on the block (draw a free body diagram!) and then apply Newton's 2nd law.

I did draw a diagram. F = mu N yeah?
 
ChEJosh said:
I did draw a diagram. F = mu N yeah?
Yep. What's the normal force? Now find the net force parallel to the ramp.

As Feldoh says, keep it simple. (No need for calculus.)
 
N = mg cos(theta)

So,
a = g sin(theta) - mu g cos(theta)?

Then,
v = v_0 +at

Since we're looking at when it comes to rest, v=0

T = -v_0/a

Look right?
 
Looks good.
 
Excellent. Thank you.

For the second part, with the axes being the vertical and horizontal, I'm going to have a multi-component acceleration, yeah?

In the y-direction, the forces working is the weight/gravity which is pointing in the -y-direction, and components of the Normal and Friction pointing in the +y-direction.
For the x-direction, I'm going to have the normal in the positive x-direction, and the friction in the -x-direction.
Does that sound right?
 
Sounds good.
 
  • #10
For part a, I got
T = -v0/(g sin(theta) - mu g cos(theta)]

Obviously since it's the same situation, I should be able to get the same answer for part b, But I'm having difficulty.

x-direction
Fx = N cos() - mu N cos()
ax = [N cos() - mu N cos()]/m
T = -[v0x cos() m]/[N cos() - mu N cos()]

y-direction
Fy = mu N sin() + N sin() - mg sin()
ay = [mu N sin() + N sin() - mg sin()]\m
T = -[v0y sin() m]/[mu N sin() + N sin() - mg sin()]

theta is in all the empty (), of course.

And then, I'm stuck. I don't see how I could get the original answer from that, so either it's something I just haven't seen or I did something wrong.
 
  • #11
ChEJosh said:
x-direction
Fx = N cos() - mu N cos()
Check those components. They can't both be cosine. :wink:
 
  • #12
Doc Al said:
Check those components. They can't both be cosine. :wink:

No? I always thought that if it was an x component it was cosine, and y component was sine. Those are both the x components, so...
 
  • #13
ChEJosh said:
No? I always thought that if it was an x component it was cosine, and y component was sine. Those are both the x components, so...
Sure, if the angle is with respect to the x-axis. But theta is the angle of the ramp. What's the angle of the normal force?
 
  • #14
Doc Al said:
Sure, if the angle is with respect to the x-axis. But theta is the angle of the ramp. What's the angle of the normal force?
wrt the ramp 90º or wrt x-axis 90º+theta

How is the angle not wrt the x-axis? It's the angle made by the ramp with the x-axis
 
  • #15
ChEJosh said:
wrt the ramp 90º or wrt x-axis 90º+theta
That's with respect with the -x-axis, but OK. (What's the angle with the +x-axis?) In any case, you wrote cos(theta).
 
  • #16
Doc Al said:
That's with respect with the -x-axis, but OK. (What's the angle with the +x-axis?) In any case, you wrote cos(theta).

How is it the -x-axis?

Although, I now see that you're saying that the normal is theta from the positive y-axis, and needs to be sin(theta), right?
 
  • #17
Right.
 
  • #18
So now I get

x-direction
Fx = N sin() - mu N cos()
ax = [N sin() - mu N cos()]/m
T = -[v0x cos() m]/[N sin() - mu N cos()]

Substitute N = mg cos()

T = -[v0x cos() m]/[mg cos() sin() - mu mg cos() cos()]

cancel m cos() to arrive at original answer, right?

Thanks for all of your help.
 
  • #19
ChEJosh said:
cancel m cos() to arrive at original answer, right?
Looks good to me. (And you're welcome.)
 
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