Homework Help: Conservation of energy/Finding maximum height

1. Oct 26, 2016

fishturtle1

1. The problem statement, all variables and given/known data
A 28-kg rock approaches the foot of a hill with a speed of 15 m/s. This hill slopes upward at a constant angle of 40.0∘ above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively.

a)Use energy conservation to find the maximum height above the foot of the hill reached by the rock.

2. Relevant equations
Conservation of energy for this problem is: PEi + KEi + Workstatic friction = PEf + KEf + Workkinetic friction

Wfriction = μmgcos(Φ)

Friction force = μmg

3. The attempt at a solution
m=28kg
vi=15m/s
Φ=40°
μs=0.75
μk=0.20

fs=0.75mgcos(Φ)
fk=0.20mgcos(Φ)

Wfs=.75mgcos(Φ)*s*cos(180)
Wfk=0.20mgcos(Φ)*s*cos(180)

plug all known variables into my equation for conservation of energy:

(.5)(28)(152)+(28)(9.8)(0)+(0.75)(28)(9.8)(cos(40))*s*(-1) = (.5)(28)(02)+(28)(9.8)(hf)+(0.20)(28)(9.8)(cos(40))*s*(-1)

which equals:

3375J - 157.65*s = 274.4*hf - 42.041*s

Then i tried to find s which is the displacement vector by doing Newton's second law

ΣF= max = wx - fk

28a = 28*9.8*sin(40°) - 0.2*28*9.8*cos(40)

28a = 134.34

a = 4.80m/s2

Then i used vf2 = v22 + 2ad, where d would be my s vector

02 = 152 + 2(4.80)d
d = -225/9.6 = -23.43 which i made positive because i just wanted the magnitude.

But then my final answer is hf = 22.18m, which isn't correct.

2. Oct 26, 2016

Simon Bridge

You have contradictory work equations ... one says $W=\mu mg \cos\theta$ and the other says $W=\mu mg \cos\theta s \cos\theta$
You say "s" is the displacement vector - but you have not defined it.
You have a statement for work against static friction ... shouldn't this be zero?

3. Oct 26, 2016

Staff: Mentor

You should consider doing more of your work symbolically rather than plugging in numbers right away. Besides being a pain for others to follow what numbers correspond to which quantities to check the logic, it often means a lot more work since many times things will cancel out along the way when symbolic expressions are simplified. Rounding calculations along the way also introduces rounding/truncation errors and can affect your significant figures as they accumulate.

I noticed that you included a term for static friction in your energy conservation equation. Static friction does no work, since it only exists when the object is not moving. So it shouldn't be part of your energy conservation considerations.

I also noticed that you abandoned the energy conservation approach after reducing the expression to an equation with s and hf. Why? You could have expressed s in terms of hf (since they're related by trig thanks to the constant slope of the hill) and solved for hf.

I suggest that you revisit your energy conservation equation. Start by spelling out the terms symbolically. You should find that the mass at least can be cancelled out.