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

In summary: I now see that you're saying that the normal is theta from the positive y-axis, and needs to be sin(theta),...Thanks for catching that! So, part a is v = v_0 + atand part b is v = v_0 - atwhich are the same, since they're looking at when the block is at rest. How is it the -x-axis? It's the same as with the y-direction. It's the angle made by the ramp with the x-axis.
  • #1
ChEJosh
27
0

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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  • #2
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.
 
  • #3
You're making it too complicated, just find the acceleration and use those old constant acceleration motion equations.
 
  • #4
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?
 
  • #5
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.)
 
  • #6
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?
 
  • #7
Looks good.
 
  • #8
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?
 
  • #9
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.)
 

1. What is friction?

Friction is a force that resists the motion of an object when it comes into contact with another object. It is caused by the irregularities and roughness of surfaces, and it always acts in the opposite direction of an object's motion.

2. How does friction affect a block on a ramp?

When a block is placed on a ramp, friction plays a crucial role in determining its movement. The force of friction acts in the opposite direction of the block's motion, causing it to slow down. This means that the steeper the ramp, the greater the friction force, and the slower the block will move.

3. What factors affect the amount of friction on a block on a ramp?

The amount of friction on a block on a ramp is affected by several factors. These include the roughness of the surfaces in contact, the weight of the block, the angle of the ramp, and the type of surface the block is sliding on.

4. Can friction be reduced on a block on a ramp?

Yes, friction can be reduced on a block on a ramp by using a smoother surface or by applying a lubricant between the block and the ramp. Increasing the angle of the ramp can also decrease the amount of friction acting on the block.

5. How does friction affect the energy of a block on a ramp?

Friction has a significant impact on the energy of a block on a ramp. It converts some of the block's kinetic energy into heat, causing the block to lose speed and eventually come to a stop. This is why friction is considered a dissipative force as it transforms mechanical energy into thermal energy.

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