What is the speed of the block sliding down an inclined plane?

[I Like Pi]

Homework Statement

A 2.5 kg wooden block slides from rest down an inclined plane that makes an angle of 30o with the horizontal. If the plane has a coefficient of kinetic friction of 0.20, what is the speed of the block after slipping a distance of 2.0 m?

Homework Equations

The Attempt at a Solution

I know how to solve this using dynamics and kinematics equations, but I'm doing a unit of energy and momentum, is it possible to solve this using momentum and/or energy equations?

thanks for your time

 

[tiny-tim]

Hi I Like Pi!

I know how to solve this using dynamics and kinematics equations, but I'm doing a unit of energy and momentum, is it possible to solve this using momentum and/or energy equations?

Yup!

Use the work-energy theorem … https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in mechanical energy.

 

[I Like Pi]

Hi I Like Pi! Yup!

Use the work-energy theorem … https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in mechanical energy.

Hey, thanks! So would I use Ek = Ep - Wf, Wf being (muN)dcosx?

would that work?

Thanks tim

 

[tiny-tim]

Hi I Like Pi!

(have a mu: µ and a theta: θ and a degree: ° )

(just got up :zzz: …)

Hey, thanks! So would I use Ek = Ep - Wf, Wf being (muN)dcosx?

Yes …

except why cosx? … the friction is parallel to the slope, not horizontal

(and don't call it W_f, it's simply W;

also we usually write KE and PE …

then we can write eg KE_i, which is a lot easier to read than E_ki ! )​

 

[I Like Pi]

Hi I Like Pi!

(have a mu: µ and a theta: θ and a degree: ° )

(just got up :zzz: …)


Yes …

except why cosx? … the friction is parallel to the slope, not horizontal

(and don't call it W_f, it's simply W;

also we usually write KE and PE …

then we can write eg KE_i, which is a lot easier to read than E_ki ! )​

Haha, thanks

Well, i used the cosθ because the only equation we are using in the energy/momentum unit for work is W = F*cosθ*d

The equation I used was:
PE = KE + W
Therefore,
KE = PE - W But you're work has to be negative because it is in the opposite direction of your potential energy, so:
.5mv^{2} = mgh - [(µ*mg*cosθ)*cosθ*d]
Plug in your values, and you get about 3.4 m/s.

The answer is supposed to be 3.5 m/s, but then again, I guess its the same thing, or you could just include the neg because you know it's supposed to be neg (opposite direction)

Thanks for your help tim!

 

[tiny-tim]

Hi I Like Pi!

(you can't write µ and θ in LaTeX … you have to write \mu and /theta )

Looks ok , except I still don't see why you have two cosθs.

(the θ in your F*cos θ*d formula is the angle between the friction F and d, but that's zero)