Forces and Motion Question

  • Thread starter BassMaster
  • Start date
  • #1
31
0

Homework Statement



A skier skiing downhill reachers the bottom of a hollow with a velocity of 20 m/s, and then coasts up a hill with a 10 degree slope (@). If the coefficient of kinetic friction is 0.10, how far up the slope will she travel before she stops?



Homework Equations



Fg=mg
Ff= uFn
Fslope = (m)(g)(sin @)
Fn = (m)(g)(cos @)
v2^2= v1^2 + 2ad


How would you solve this?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
I would use conservation of energy. The initial kinetic energy of the skier is equal to the sum of the gravitational potential energy she gains going up the hill plus the work done in overcoming the frictional force travelling up the hill. Can you write down an expression for each of these three energies?
 
  • #3
31
0
I would use conservation of energy. The initial kinetic energy of the skier is equal to the sum of the gravitational potential energy she gains going up the hill plus the work done in overcoming the frictional force travelling up the hill. Can you write down an expression for each of these three energies?
This question was written assuming we know nothing about the law of conservation of energy. There must be a way to solve it by using only equations related to forces and motion.
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
Ok. So force acting on the skier parallel to the slope is your Ff+Fslope. That should let you find the acceleration parallel to the slope. Put that into your last equation with a final velocity of zero.
 
  • #5
31
0
The problem is that NO MASS is given. So I can't solve for any of those values.
 
  • #6
960
0
don't need mass as it cancels out. say the slope is 0,

ma=N*mu =mg*mu a=g*mu
 
  • #7
31
0
I'm sorry I didn't quite understand what you wrote there.
 
  • #8
960
0
just that the mass cancels. sum forces=ma=Normal force*frictional coefficient
Normal=-mg so a=-g*friction coeff. this is for a slope of 0, but the masses will all cancel in any event.
 

Related Threads on Forces and Motion Question

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
906
  • Last Post
Replies
0
Views
886
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
391
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
Top