Solving a Block's Motion on Inclined & Horizontal Surfaces

[leighzer]

A block of mass 3.0 kg starts at a height 0.60 m on a plane that has an angle of 30 degrees from the horizontal. Upon reaching the bottom, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is 0.20, how far does the block slide on the horizontal surface before coming to rest?

I found the net force on the block and from there the acceleration, but i think this problem is supposed to be solved using Conservation of Energy theory, or something along those lines.

I know E initial cannot = E final since friction is acting on the block, but I'm not sure how to work that into the problem.

Can anyone help with this question?

 

[leighzer]

If I can find the speed of the block at the bottom of the inclined plane, I'll be able to work it out from there, but how do i find the speed when friction is present?

 

[neutrino]

If you take the block, ramp, floor and the air surrounding all these, the energy will be conserved. But, in this problem, only the block is important. So, while the block's coming down, it gains kinetic energy, but also loses some energy due to friction. You know how much it had initially, this should tell you how much it would have when it is at the bottom of the ramp. The second part is quite simple once you know at what speed the block leaves the ramp.

 

[leighzer]

So are you saying the energy at the top will equal the energy at the bottom of the ramp?

 

[neutrino]

If that were so, then the block would not have lost energy.

Energy(top) = Energy(bottom) + Energy(lost)

 

[leighzer]

Right, but how do you figure out how much energy is lost due to friction? Is it something to do with Work?

 

[neutrino]

Right, but how do you figure out how much energy is lost due to friction? Is it something to do with Work?

Exactly. The plane does work on the block. Now, if you know the force due to friction and the definition of work, finding the work done should be simple. :)

 

[leighzer]

Ok, thanks for the help.