A Car Travels Down a Ramp and Continues Traveling After Leaving the Ramp but Eventually Stops

You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.

• Forums
• Homework Help
• Introductory Physics Homework Help

How far a block travels after sliding down a ramp

• Thread starter Daedris
• Start date Mar 4, 2009

Homework Statement

A block starts from rest at a height of 3.1 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s2 .

If the block continues to slide on the ground with the same coefficient of friction, how far
will the block slide on the ground until coming to rest?

µ = .22
θ = 25° (angle of the ramp's incline)
Height of block on ramp = 3.1 meters
mass of block = 6.6kg

Homework Equations

ΣF = ma
Position(time) = Position initial + Velocity initial X time + .5(accelaration)(time)

The Attempt at a Solution

The first part of this problem is to determine the speed of the block at the bottom of the ramp. Not going into too much detail as to how I solved that, I determined that the blocks speed at the bottom of the ramp is 5.662 m/s, and that the answer is correct. I also found that the acceleration of the block is 2.187 m/s^2.

The part I cannot figure out is how far the block will slide. I started by saying:
accelartion in angled direction = 2.187 m/s^2
acceleration in horizontal direction = 2.187cos(25) = 1.982 m/s^2

Velocity in angled direction = 5.662 m/s
Velocity in horizontal direction = 5.662cos(25) = 5.132 m/s

I know when the block is on the ground, it has 4 forces: Sliding force of the block, Force of Kinetic Friction, mass*gravity, and normal force. Since there is no incline, normal force is equal to mass*gravity.

Force of kinetic friction = µ*normal force = (.22)(9.8)(6.6) = 14.23 N
acceleration of kinetic friction = µ*gravity = (.22)(9.8) = 2.156

acceleration of the block = acceleration in the horizontal direction = 1.982 m/s^2 (from above). Since acceleration of kinetic friction is opposite the direction of the block's acceleration, I subtract those 2 values: acceleration = 1.982 - 2.156 = -.174 m/s^2.

Using the kinematic equation listed above I see that:

Distance = Initial pos + Initial Velocity*time + .5*acceleration*time^2 = 0 + 5.132*time - .087*time^2

Velocity = 0 = Initial velocity + acceleration*time = 5.132 -.174t => t = 29.5 seconds

Distance = 0 + 5.132(29.5) - .087(870.25) = 151.394 -75.712 = 75.682 meters

I do not think I am doing this correctly. Can anyone point me in the right direction?

Answers and Replies

Looks like there are two difficulties here.
1) When the block moves out onto the horizontal plane, the previous component of acceleration from moving down the plane no longer applies. You need a new FBD for the horizontal motion.
2) You don't appear to accounted for the friction on the horizontal plane correctly. Where do you get this mu*g term? There is no such thing as an "accel of kinetic friction."

You need an FBD, and remember that it contains only forces, no m*a terms!

ok simply put you need to do a suvat equation cant rember which one that will work out s and it must contain v then put v=0 and fill in the other numbers then you have your distance as whatever s=

Looks like there are two difficulties here.
1) When the block moves out onto the horizontal plane, the previous component of acceleration from moving down the plane no longer applies. You need a new FBD for the horizontal motion.
2) You don't appear to accounted for the friction on the horizontal plane in your work.

Thanks for the reply, but I think I did both of those. I said that Friction is equal to (.22) which is the coefficient of friction times the normal force. If I divide by mass, I get the acceleration in the x-direction due to friction, which is .22g = 2.156 m/s^2.

I drew a new block diagram for when the block is at the bottom of the ramp. I used the acceleration in the x-direction at the time the block is at the bottom of the ramp, which I got to be 1.982 m/s^2.

I assume to get the acceleration of the block in the positive direction x, I take the x-component of the acceleration, 1.982, minus the acceleration due to friction, 2.156. That gives me -.174.

I have thought the problem through and still am not doing something correctly.

Thanks for any help.

Edit: For the kinetic acceleration, I said that kinetic force is a retarding force in the opposite direction of the sliding force of the block. If I know that the Kinetic Force is equal to coefficient of friction*mass*gravity, then I can divide by the mass to get the acceleration effect of the kinetic force, correct?

I have no way to know what you are doing because you are not working with equations. You are simply calculating various terms, one by one. If you want help, you will need to write a complete equation, a statement of equality. The perform some operation and write the equation again. Do this over and over until you have a solution. It is the only way your work can be followed, and the only way you will be able to follow it next week.

Welcome to PF.

This problem is a little easier to consider from a Potential Energy/Kinetic energy POV.
PE - W_fr = KE

m*g*h - μ*m*g*cos25*h/sin25 = m*g*h - μ*m*g*cotan25*h = ½mv²

v² = 2*g*h*(1 - μ*cotan 25)

This yields your v = 5.662 m/s directly.

This means also that you have ½mv² of kinetic energy at the bottom = 105.8 J.

Frictional work robs that energy at μ*m*g*X

So X = 105.8/μ*m*g

If you are determined to work it with F = m*a then you can use your V and recognize that acceleration on the flat = - μ*g

V² = 2*a*X or

X = V²/(2*μ*g)

Congratulation, LowlyPion! You have just wiped out the learning opportunity for the student. What is the point for that?

Welcome to PF.

This problem is a little easier to consider from a Potential Energy/Kinetic energy POV.
PE - W_fr = KE

m*g*h - μ*m*g*cos25*h/sin25 = m*g*h - μ*m*g*cotan25*h = ½mv²

v² = 2*g*h*(1 - μ*cotan 25)

This yields your v = 5.662 m/s directly.

This means also that you have ½mv² of kinetic energy at the bottom = 105.8 J.

Frictional work robs that energy at μ*m*g*X

So X = 105.8/μ*m*g

If you are determined to work it with F = m*a then you can use your V and recognize that acceleration on the flat = - μ*g

V² = 2*a*X or

X = V²/(2*μ*g)

Thanks very much! Using what you said that there is no acceleration from the block when it is on the flat surface, I figured out that the distance will be 7.434 meters and that is the correct answer! Thanks very much!

However, I do not fully understand: Why does the block lose its acceleration when it gets to the flat ground? The only acceleration is kinectic friction, but why is that?

Again, thank you very much!

Edit: I am sticking with kinematic equations as I have not learned much of anything about Work yet.

Congratulation, LowlyPion! You have just wiped out the learning opportunity for the student. What is the point for that?

Wrong, "Dr. D". I had already worked the problem and I was just hung up because I thought the block maintained acceleration. Thanks to LowlyPion, I DID learn something.

However, I do not fully understand: Why does the block lose its acceleration when it gets to the flat ground? The only acceleration is kinectic friction, but why is that?

Kinetic Friction is a force that resists motion. And of course it is present the whole way. On level ground then it acts as a decelerating force against its motion.

F = m*a = - μ*m*g

This means that your a = - μ*g

Kinetic Friction is a force that resists motion. And of course it is present the whole way. On level ground then it acts as a decelerating force against its motion.

F = m*a = - μ*m*g

This means that your a = - μ*g

Thanks very much for your help!

I have a problem with the solution presented by LowlyPion.

The block's velocity at the end of the ramp is v = 5.662 m/s. Then the block enters the horizontal part of the course on the ground. But this involves a change in the direction of the velocity. The component of the velocity perpendicular to the ground just before the block entered the ground was v sin(25°). Then, when it starts to move on the ground, it is zero.

This means that the integral of the vertical component of the normal force over time from when the block is at the end of the ramp to just after equals minus M v sin(25°). Here we assume that the transition from the ramp to the ground happens over a very short distance so that we can ignore gravity here.

So, you have a huge normal force that acts on the block, leading to the change in momentum in the vertical direction, therefore there must be a huge friction force leading to a change in the velocity in the horizontal direction as well. This is then of the order of mu times v sin(25°), but the precise value depends on the way the transition happens.

Anyway, what is clear is that it is wrong to take the velocity of the block when it starts to move on the ground to be 5.662 m/s.

Anyway, what is clear is that it is wrong to take the velocity of the block when it starts to move on the ground to be 5.662 m/s.

What would you propose for initial velocity then on the flat part?

ok what you need to do is look at it as two separate cases. one in which the block is on the slope and one which it is not. the force of friction should affect the block's motion a bit differently depending on whether it is on the slope or not so you should find the velocity of the block at the time that it reaches the bottom of the slope then again when it is off the slope for as stated before, the force of friction will be different even thou the friction coefficient is the same. gimme a sec i will do the calculations and give you the answer.

ok i think the V at the bottom of the ramp should actually be more like 16.04697354m/s and we are assuming that all of the energy is transferred over so that means that at the beginning of the plain the block is going that fast. now we use V2^2 = V1^2 + ad and V2 is gonna = 0 because we want to know when it stops. a is going to be -2.156 because these are vectors we are talking about and the acceleration is opposite the motion. we have V1 so its simply V1^2/a = d = 119.4366233m. remember, once gravity is out of the picture i.e. the box is on flat ground, the only acceleration is that caused by the force of friction. oh and by the way, your velocity at the bottom of the ramp is wrong... like waaaaaaaaaaay wrong. its simply not possible. remember, its accelerating over the distance of the hypotenuse of the triangle. also, double check my answer if you want, i sometimes make stupid calculator errors lol

What would you propose for initial velocity then on the flat part?

I had a dream last night. In that dream I answered this question here, but unfortunately I don't see that posting now. So, I'll now have to type it again

During the rapid change of direction, we can ignore gravity. We then have:

|tangential acceleration| = mu times |acceleration in normal direction|

Tangential acceleration = d|v|/dt

acceleration in normal direction = |v| d theta/dt

We thus have:

V_flat = V_ramp*exp(-0.22*25°) = V_ramp*exp(-0.096)=

0.9 V_ramp = 5.1 m/s

So, this means that the seemingly insignificant change of direction will lead to the block traveling only 80% of the predicted distance.

2.187m/s^2 is the acceleration of the block down the slope with friction taken into acount so just use algebra to get the distance of the slope then use V2^2=V1^2*ad... i just reolised that my answer that i gave in my previous post was wrong lol oops. the velocity at the botom of the ramp should be more like 4.005263646m/s shouldnt it?

Suggested for: How far a block travels after sliding down a ramp

• Last Post
• Apr 3, 2022

• Last Post
• Mar 23, 2022

• Last Post
• Oct 5, 2022

• Last Post
• Aug 20, 2021

• Last Post
• Apr 18, 2021

• Last Post
• Apr 14, 2022

• Last Post
• Oct 3, 2021

• Last Post
• Mar 22, 2022

• Last Post
• Nov 24, 2020

• Last Post
• Aug 5, 2022

• Forums
• Homework Help
• Introductory Physics Homework Help

thrasheroultiventy.blogspot.com

Source: https://www.physicsforums.com/threads/how-far-a-block-travels-after-sliding-down-a-ramp.297204/

Share this post