Solve Work-Energy Problem: 2kg Mass Down Frictionless Hill

[sasuke07]

Homework Statement

In the illustrated setting below a 2 kg mass slides down a frictionless hill starting at rest from a location 5 m above the floor. When the mass reaches the floor it slides across the floor and up another frictionless ramp on the far side. The horizontal floor has a coefficient of friction of 0.2 which is observed to remove 30% of the kinetic energy from the mass as it slides across the unknown width of the floor.

I am aware this is a fairly contrived setting and it was tempting to draw a specific example from the Honda ad we investigated in lab. Because this is pretty new material for most of you I wanted to limit the distraction and confusion of surrounding objects. That doesn't mean there aren't any distractions in this problem so be careful:)

Homework Equations

PE=mgh
Ke=.5mv^2
inital Ke+initial Pe=final KE +final PE

The Attempt at a Solution

0+mgh=.5mv^2+0
v=SQroot of 2gh
v=10m/s
Inital PE is 100
final KE is 100 using the formulas from above.

so my question is what else i could figure out given the problems and how i would figure it out. Thanks.

 

[PeterO]

Homework Statement

In the illustrated setting below a 2 kg mass slides down a frictionless hill starting at rest from a location 5 m above the floor. When the mass reaches the floor it slides across the floor and up another frictionless ramp on the far side. The horizontal floor has a coefficient of friction of 0.2 which is observed to remove 30% of the kinetic energy from the mass as it slides across the unknown width of the floor.

I am aware this is a fairly contrived setting and it was tempting to draw a specific example from the Honda ad we investigated in lab. Because this is pretty new material for most of you I wanted to limit the distraction and confusion of surrounding objects. That doesn't mean there aren't any distractions in this problem so be careful:)

Homework Equations

PE=mgh
Ke=.5mv^2
inital Ke+initial Pe=final KE +final PE

The Attempt at a Solution

0+mgh=.5mv^2+0
v=SQroot of 2gh
v=10m/s
Inital PE is 100
final KE is 100 using the formulas from above.

so my question is what else i could figure out given the problems and how i would figure it out. Thanks.

It would seem to me that on the way down the frictionless slope, 100% of the initial PE will be converted to KE.
On the way across the floor, 30% of that energy is lost to friction.
On the way up the other side, the remaining 70% is converted back to PE as it climbs the frictionless slope.

There is my conceptualisation for you to work with.

 

[sasuke07]

Thanks for the reply, So i still need to answer to quesitons in regards to this problem and that's how far the box will travel without the second slope and how far up the box will travel up the second slope. I don't really know where i would begin in regards to those last 2 problems

 

[LawrenceC]

For the first question, what do you know about the relationship between work and energy? For the second question PeterO has already supplied you with a very good hint.