Conservation of energy and GPE problem

[mmattson07]

Homework Statement

A 2.7 kg bundle starts up a 27° incline with 106 J of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is 0.42?

Homework Equations

K= 1/2mv^2
conservation of energy => Ki + Ui = Kf+Uf

The Attempt at a Solution

Initially the gravitational potential energy is 0 (y=0) => initial Emec=106J
but how does the friction come into play in this problem? Does the coeff of kinetic friction times the distance give you the energy transferred to thermal heat? If some one could walk me through this...

 

[mmattson07]

Correction the energy is NOT conserved due to the friction force between the bundle and incline so I'm guessing the final equation will have some -thermal energy?

 

[Delzac]

You start off with kinetic energy.

You end off with gravitational potential energy and energy lost due to friction.

delzac

 

[mmattson07]

Don't you have to know how far it will slide up before you can calculate the final gravitational potential energy? Same with the energy lost?

 

[Delzac]

You will have to.

But it is basically one equation one unknown.

 

[mmattson07]

Ok, so we have

106J+0J=mgdsin27-(0.42)d
=>106J= d(mgsin27-0.42)
=>d=106J/(mgsin27-0.42)

Just realized: we can assume Kf=0 because the bundle will be stopped?

 

[Delzac]

Yup, that's correct.

 

[mmattson07]

Not getting the correct answer.

d= 106J/((2.7kg)(9.8m/s^2)sin27-0.42)
`=9.144 m

The correct answer is 4.837m

 

[mmattson07]

I'm stumped. Here is what the hint reads: "You then need to set up an energy equation: initial mechanical energy - energy transferred to thermal energy = final mechanical energy."

So I have been trying:

106J-(mg/cos27)(0.42)d=mgdsin27
=>d=106J/(mgsin27+(mg/cos27)(0.42))

with mg/cos27 being the normal force...but I'm still missing something.

 

[mmattson07]

oh NVM