What is the Speed of a Book Sliding Down an Inclined Plane?

[The_big_dill]

Homework Statement

A Book (initially at rest) slides down an inclined plane, when it is at the bottom of the slope, its speed is 2.5 m/s. What is its speed half way and a third way down the slope.

V₁ = 0
t₁ = 0
V₂ = 2.5 m/s

Homework Equations

a = \Deltav / \Delta t

v = \Deltad / \Delta t

Any of the other Uniform motion equations (kinematics)

The Attempt at a Solution

It looks simple, but i simply cannot figure out where to start...

I have to many missing variables, which unless i can assume something or equate 2 equations, i have no idea what to do.

I know for a fact that i am not suppose to be using forces (ex. x-component of acceleration due to gravity), i should be using kinematics.

 

[tiny-tim]

Hi The_big_dill!

(have a delta: ∆ )

A Book (initially at rest) slides down an inclined plane, when it is at the bottom of the slope, its speed is 2.5 m/s. What is its speed half way and a third way down the slope.

Use the standard constant acceleration equations.

You have vi vf and s, and you want to find vf for a different s …

what equations do you think you should use? ​

 

[Ambidext]

Also, becareful which axis you use.

If you want to take the book's direction of travel to be the x-axis, then you need to ressolve for g.

If you wish to take g as the y-axis, then you need to ressolve for the x and y components of the sliding motion.

After you've decided on your axis, apply your equations of motion as suggested ^^

 

[tiny-tim]

Hi Ambidext!

No, the question doesn't give the angle.

This is a purely 1D problem, not needing g, in which we could work out the angle from the given data, but we're not asked for it.

 

[The_big_dill]

Hi Ambidext!

No, the question doesn't give the angle.

This is a purely 1D problem, not needing g, in which we could work out the angle from the given data, but we're not asked for it.

Exactly, thank you for clarifying that XD

Also, tiny tim, i am either not familiar with that equation, or i already listed it in my original post.

If you mean a = \DeltaV/\Deltat, then i don't have \Deltat, i only have t1... which is 0... which is undefined.

 

[tiny-tim]

(What happened to that ∆I gave you? )

There are three standard constant acceleration equations … look them up.

Then call the distance "s" (and later s/2 and s/3), and carry on from there.

 

[The_big_dill]

I think there is some confusion here...

I use "d" to represent distance

V₁ to represent initial velocity

V₂ to represent final velocity

The only other equation that makes sense to me (but still missing multiple variables):

V₂² = V₁² + 2a\Deltad

I don't have distance, i also don't have acceleration...

 

[tiny-tim]

ok, then call the distance "d" (and later d/2 and d/3), and carry on from there

 

[The_big_dill]

ok, then call the distance "d" (and later d/2 and d/3), and carry on from there

I cannot solve for d. I do not have 'a', or delta 't'.

Maybe i am missing something, but i do not see it.

 

[tiny-tim]

What equations do you know involving d and t and only one v ?

(If you don't know any, look it up)

 

[The_big_dill]

What equations do you know involving d and t and only one v ?

(If you don't know any, look it up)

I do know 2 that are very similar.

\Deltad = V₁\Deltat + 1/2a\Deltat²

\Deltad = V₂\Deltat - 1/2a\Deltat²

[STRIKE]Hmmm, are you suggesting to equate them?
[/STRIKE]
[STRIKE]I should end up with a quadratic.[/STRIKE]

 

[tiny-tim]

oops!

Sorry, I forgot what the question was …

we're looking for v_f, not t, so we need an equation involving d and two vs (but no t).

 

[The_big_dill]

Sorry, I forgot what the question was …

we're looking for v_f, not t, so we need an equation involving d and two vs (but no t).

V₂² = V₁² + 2ad

 

[tiny-tim]

Yes.

Go on, then. ​

 

[The_big_dill]

Yes.

Go on, then. ​

... I am missing 2 variables, d and a.

2.5² = 0² + 2a\Deltad

 

[The_big_dill]

For Δd/2, the equation becomes,

v²² = 2aΔd/2 ...(2)

Divide eq.(1) by (2), and solve for v₂

Interesting... I will give that a go.

I already know V₂.

 

[rl.bhat]

V₂² = V₁² + 2ad...(1)

Write two more equations for d/2 and d/3.
To eliminate unknown quantities, divide eq.(1) by eq.(2) and (3) and solve for required velocities.

 

[SammyS]

What's the problem with just using conservation of Energy.?

 

[The_big_dill]

Write two more equations for d/2 and d/3.
To eliminate unknown quantities, divide eq.(1) by eq.(2) and (3) and solve for required velocities.

I already have velocity 1 and 2... i need distance, time or acceleration.

 

[The_big_dill]

What's the problem with just using conservation of Energy.?

I do not know how you are going to solve it without any angles.

 

[SammyS]

I do not know how you are going to solve it without any angles.

You just need the relative heights. I assume the incline has a constant slope!

So at half way down the incline, h_1/2 =(1/2) h, where h is the initial height.

And h_1/3 =(2/3) h.

mgh‒mgh_(1/2) = (1/2)mgh.

mgh‒mgh_(1/3) = (1/3)mgh.

From the given information, mgh = (1/2)m(2.5)^2, in units of Joules.

Added in Edit:

Well, I should have accounted for friction. The energy dissipated by friction is proportional to the distance down the incline which is proportional to Δh, so it works out the same.

Looks like nobody else has even mentioned friction.