# Box Sliding Up a Ramp

## 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

$$x = x_0 + v_0 t + (1/2) a t^2$$

$$v = v_0 + a t$$

$$v^2 = v_0^2 + 2 a \Delta x$$

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

## The Attempt at a Solution

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

as

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a} = -mgsin37 - .30mg + F_t$$

With $$F_t$$ 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

$$v = v_0 + a t$$

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.

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## 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?

oops... fixed it.

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.