Please help me on this FORCES problem

  • #1
emeraldempres
30
0
A skier of mass 61.0 kg starts from rest at the top of a ski slope of height 61.0 m.
a)If frictional forces do −1.04×104 J of work on her as she descends, how fast is she going at the bottom of the slope?

b)Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is = 0.180. If the patch is of width 68.0 m and the average force of air resistance on the skier is 180 N, how fast is she going after crossing the patch?

c)After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 3.00 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

I have only started attacking part a and i looked at the other parts and do not know where to start.

on part a i started by using the eqquation work = .5 (mass)(velocity^2) but my answer was wrong.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,465
1,956
Use energy methods. What's her starting mechanical energy? How much of that energy is "lost" to friction? How much is left when she gets to the bottom of the hill?
 
  • #3
konthelion
238
0
By the work-energy theorem, [tex]0=\Delta E_{mech}+\Delta E_{therm}[/tex]
Hence, by for problems with kinetic friction,
[tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex] where [tex]E_{mech}=\Delta U+\Delta K[/tex]

where U = potential energy, K = kinetic energy, and therm is the energy lost. You will then be able to find [tex]K_{f}=\frac{1}{2}mv_{i}^2[/tex]
 
  • #4
emeraldempres
30
0
in trying this, i tried to find the potential energy which i found to be 36465.8 N by using u= mgy (m=61 kg, g= 9.8 meters per second squared, y= 61 m). in doing that the only unknown i had was velocity and it came out to be 34.57 m/s and that was wrong.. did i mess up?
 
  • #5
emeraldempres
30
0
her starting mechanical energy is 0 right?
 
  • #6
emeraldempres
30
0
she lost -1.04*10^4 J to friction
 
  • #7
emeraldempres
30
0
and i do not know how to calulate how much is left at the bottom
 
  • #8
konthelion
238
0
[tex]\Delat U = K_{f}+\Delta E_{therm}[/tex] Ok, so you know that [tex]E_{therm}=-1.04*10^4 J[/tex] and you just calculated U. Now, find [tex]K_{f}[/tex]. Then, solve for [tex]v_{f}[/tex]
 
  • #9
emeraldempres
30
0
thank you. i got the answer
this is what i did
36465.8=.5*61v^2+1.04*10^4
26065=.5*61*v^2
854.61=v^2
v=29.2
 
  • #10
emeraldempres
30
0
on part b i find the kinetic force to be 107.6 N and the kinetic energy to 26005.52 N. am i correct in thinking that the kinectic force is going to be energy that gets converted into thermal energy?
 
  • #11
Doc Al
Mentor
45,465
1,956
on part b i find the kinetic force to be 107.6 N
That's the friction force. Don't forget air resistance.
and the kinetic energy to 26005.52 N.
That's the initial KE.
am i correct in thinking that the kinectic force is going to be energy that gets converted into thermal energy?
The work done against the combined force of friction and air resistance will equal the mechanical energy that gets converted to thermal energy.
 

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