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serenitymantra · Jun 28, 2009, 11:33 AM

A 25 kg box is initially at rest slides down a 2.0 m incline that is 15 degrees from the horizontal w/ a constant kinetic friction of 10N between the box and incline. What KE would the box have at the bottom of the incline?

I wanted to use the KE formula, but v is not given so I am stuck... can you please help? Thanks!

Perito · Jun 28, 2009, 11:40 AM

You are correct that you have to use the kinetic equation

$E=\frac 12 mv^2$

You know that the force being imposed on the box is the acceleration due to gravity. You know the mass of the box so the force is

$F=ma$

All of the force would end up as kinetic energy except for the frictional losses. The kinetic energy would be calculated just as if you dropped a ball from a given height (potential energy, E=mgh). Calculate the frictional losses and subtract that from the kinetic energy you would otherwise calculate.

serenitymantra · Jun 28, 2009, 04:15 PM

What do I do with the degrees that it is inclined at?

Originally Posted by Perito

You are correct that you have to use the kinetic equation

$E=\frac 12 mv^2$

You know that the force being imposed on the box is the acceleration due to gravity. You know the mass of the box so the force is

$F=ma$

All of the force would end up as kinetic energy except for the frictional losses. The kinetic energy would be calculated just as if you dropped a ball from a given height (potential energy, E=mgh). Calculate the frictional losses and subtract that from the kinetic energy you would otherwise calculate.

Perito · Jun 28, 2009, 05:22 PM

The potential energy is simply E=mgh which is the vertical component of the slope. The height and the mass determine the potential energy. For the normal force, you need to separate the normal force into its horizontal and vertical component. Only the vertical component is subtracted from the potential energy to get the kinetic energy.