Dynamics 1: Motion Along a Line

[Rob123456789]

<< Mentor Note -- poster has been reminded to fill out the Attempt at the Solution part of the Template in future HH threads >>

1. Homework Statement
An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope if given a very small push to get it started.

Find an expression for the object's speed at the bottom of the slope.

Express your answer in terms of the variables m, μk, L, h, and appropriate constants.

No clue how to do this.

 

[PeroK]

Homework Statement

An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope if given a very small push to get it started.

Find an expression for the object's speed at the bottom of the slope.

Express your answer in terms of the variables m, μk, L, h, and appropriate constants.

No clue how to do this.

Unfortunately, you need to have some idea how to tackle the problem and show us how far you can get on your own before you get stuck. What do you know about motion, acceleration, gravity, kinetic energy etc.?

 

[Rob123456789]

This is what I get :
final KE = initial PE - work done
½mv² = mgh - µk*mgLcos(arcsin(h/L)) → multiply by 2/m
v² = 2gh - 2µk*mgLcos(arcsin(h/L)) → take square root

v = √(2gh - 2µk*gLcos(arcsin(h/L))

 

[PeroK]

This is what I get :
final KE = initial PE - work done
½mv² = mgh - µk*mgLcos(arcsin(h/L)) → multiply by 2/m
v² = 2gh - 2µk*mgLcos(arcsin(h/L)) → take square root

v = √(2gh - 2µk*gLcos(arcsin(h/L))

You may want to do something about $\cos(\arcsin(h/L))$ but otherwise that looks right.

 

[Rob123456789]

What do I do with it ?

 

[PeroK]

What do I do with it ?

You could get rid of the trig functions.

 

[Rob123456789]

v = √(2gh - 2µk*gLcos(h/L)) ?

 

[Rob123456789]

or v = √(2gh - 2µk*gL(h/L))?

 

[PeroK]

or v = √(2gh - 2µk*gL(h/L))?

That makes no sense.

 

[Rob123456789]

Still gives me : The correct answer does not depend on: μκ.

 

[PeroK]

What do I do with it ?

$\cos$ and $\sin$ are related, which allows you to express the answer in terms of $h$ and $L$ without any trig functions. You don't have to do this.

 

[Rob123456789]

Don't really understand what you mean by expressing it in terms of h and L. This is the part I am stuck on.

 

[PeroK]

Don't really understand what you mean by expressing it in terms of h and L. This is the part I am stuck on.

You can either use $\cos^2 + \sin^2 = 1$ or go back to your diagram and express $\cos$ in terms of $h$ and $L$ using Pythagoras.