Finding the Speed Needed to Propel Packages Up an Inclined Ramp

[robbondo]

Homework Statement

You are working for a shipping company. Your job is to stand at the bottom of a 8.0-m-long ramp that is inclined at 37 degrees above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is .30 What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp?

Homework Equations

[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

[tex]v = v_0 + a t[/tex]

[tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

[tex]\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/tex]

The Attempt at a Solution

Well first I drew a digram which gave me a component of [tex]nsin37[/tex] as the force acting down the ramp and then of course the nu force of friction. SO, then I wrote out that

as

[tex]\vec{F}_{net} = \Sigma \vec{F} = m \vec{a} = -mgsin37 - .30mg + F_t[/tex]

With [tex] F_t [/tex] being the force of the throw

Then I know that since I'm solving for the initial velocity that should be related to the acceleration and the force through

[tex]v = v_0 + a t[/tex]

but I'm not sure how. I haven't been given the value for the mass, so I'm not sure if I can even solve this problem only using the given variables.

 

[radou]

Homework Statement

You are working for a shipping company. Your job is to stand at the bottom of a 8.0-m-long ramp that is inclined at 37 degrees above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is u.

Ok, and where's the problem in the text?

 

[robbondo]

oops... fixed it.

 

[denverdoc]

Am I missing something, just below where it is lopped off in the quote. was there an edit?

Rob, anyway its a bit like the last one, the masses all divide out.