• Support PF! Buy your school textbooks, materials and every day products Here!

Calculating the Work done from an Incline with Friction Problem

  • Thread starter Iconic
  • Start date
  • #1
8
0
Problem:
A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.6m and θ = 15°. The total mass of the sled and the girl is 35 kg and the coefficient of kinetic friction between the sled runners and the snow is 0.20. If the sled moves up the hill with a constant velocity, how much work is done by the father in moving it from the bottom to the top of the hill?

wG8RI2Z.png


Equations Used:

Forcenet Y components ƩFy= N-mgcosθ=0-(due to constant velocity)
So...
Normal Force Fn= mgcosθ
And since...
Friction Force Ff= (μ)kFn
So...
Friction Force Ff= (μ)k(mgcosθ)

Forcenet X components ƩFx= -mgsinθ-(μ)k(mgcosθ)

The Attempt at a Solution



Before I go any further with solving this problem I need to know how to use the applied Force from the father in this problem. I'm 100% certain my ƩFx is incorrect and was hoping someone could explain how I should fix this.
 
Last edited:

Answers and Replies

  • #2
199
15
Before I go any further with solving this problem I need to know how to use the applied Force from the father in this problem. I'm 100% certain my ƩFx is incorrect and was hoping someone could explain how I should fix this.
Yes, ##ƩF_{x}## is incorrect because you calculated Normal force incorrectly. Since ##F## (force by father) is acting horizontally, you'll have to take component perpendicular to the plane into account while calculating ##ƩF_{y}##
 

Related Threads on Calculating the Work done from an Incline with Friction Problem

  • Last Post
Replies
8
Views
6K
Replies
6
Views
5K
Replies
3
Views
10K
Replies
3
Views
2K
Replies
6
Views
11K
Replies
2
Views
13K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
11
Views
60K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
27K
Top