## friction and a crate

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 degress 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 mu_k=0.30.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?

I think im having trouble finding the normal force to this problem but im also unsure as to how to calculate the force when no mass is given...im sure it cancels somehow but just dont see it. any help would be appreciated.
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 Admin It would help to draw a vector diagram. Gravity is always acting down, and the force is given by the weight of an object (W = mg). Now friction acts parallel to a surface, but it is proportional (by mu_k) to the normal forcing acting on the surface. That force is at an angle with respect to the weight (force of gravity acting vertically). Now one can apply conservation of energy. At the beginning of the ramp, the consider the energy to be kinetic energy (KE = 1/2 mv2), and one is trying to find v. Now what happens to that KE of the box? Part of the kinetic energy goes into increasing the gravitational potential energy of the box (mgh), and one finds h from the change in elevation of an 8 m ramp inclined at 37°. The other part of the energy used in overcoming friction which acts opposite to the direction of travel, and that energy is simply Force * distance. That force as shown above is proportional to the weight (mg). So write the equation of energy balance and note that m divides out.
 im sure that this works but im not familiar with the workings of kinetic and potential energy to the point where i could solve it using this... i have only used the second law that is f=ma and that force of friction is equal to mu*the normal force. thus far i have been working this problem for what seems like an eternity and seem no closer to finding the normal force because of the lack of mass i know it is equal to cos 37 *mg and the frictional force is equal to mu times the normal force.

## friction and a crate

 Quote by trajan22 im sure that this works but im not familiar with the workings of kinetic and potential energy to the point where i could solve it using this... i have only used the second law that is f=ma and that force of friction is equal to mu*the normal force. thus far i have been working this problem for what seems like an eternity and seem no closer to finding the normal force because of the lack of mass i know it is equal to cos 37 *mg and the frictional force is equal to mu times the normal force.
You can let the mass as a parameter of the friction.
$$F_{frict} = km$$
This same force will decelerate the mass, so when you calculate the acceleration yo cancel the mass.
 so then if the force of friction is equal to cos37 *g this can also be considered the deceleration that friction provides so would that mean that gravity will also decelerate it acoriding to sin37 *9.8 then you could add the two decelerations?? if this is the case then v^2=2(a)8
 Actually, since the weight is mg, the friction is m.g.cos(37). But the acceleration is F/m, so m cancels and you have a = g.cos(37).
 so if a is equal to that i put it in this equation v^2=v_0^2 +2ax and am getting the velocity needed to be 11.19 m/s is this being approached correctly

 Quote by trajan22 so if a is equal to that i put it in this equation v^2=v_0^2 +2ax and am getting the velocity needed to be 11.19 m/s is this being approached correctly
I didn't make the calculation, but the reasoning is correct.