MCAT Physics Question: Mechanics III

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Homework Statement



Basically, Alice is pushing her brother Jeff on a toboggan at an ice skating rink that is flat except for a ramp that has an incline of 30 degrees

Alice weighs: 60 kg
Jeff weighs: 28 kg
Toboggan weighs: 2 kg

They decide to push Jeff and the toboggan up the ramp. They start from rest 10 m from the incline and she pushes him with a force that varies with the distance [the book has a figure of Force (N) vs. Distance (m)]. You can see the diagram if you google the question below.

http://books.google.ca/books?id=HCU... she lets go just before the incline?&f=false

Jeff goes speeding up the incline with a velocity of 2 m/s

The Question is:

"How much work does Alice do from the moment she begins to push Jeff 10 m from the incline until she lets go just before the incline?"

A) 60J
B) 75J
C) 0J
D) There is not enough information to answer the question


Homework Equations



I know how to solve the problem, I realize Figure 1: Force vs. Distance graph will give me the amount of work applied if I add the area under the graph. The problem is the graph doesn't show the Force and I am not sure how to solve for it.

F=ma

Vf^2=Vi^2 +2ad

a= Vf^2/d



The Attempt at a Solution




I figure the mass is the Toboggan + Jeff.

But solving for acceleration is the problem I am having. I know the initial velocity will equal to zero, but the velocity during the distance traveled I cannot figure out

The answer they give for Force = 7.5 N


Any help would be great,

Thank you
 

Answers and Replies

  • #2
I'm not any expert but I think using the Work Energy Theorem will quickly give you an answer for work. I think your mass of Jeff + Toboggan was correct.

Work = KE[itex]_{f}[/itex]-KE[itex]_{i}[/itex] = [itex]\frac{1}{2}[/itex]mv[itex]_{f}[/itex][itex]^{2}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]_{i}[/itex][itex]^{2}[/itex]

Hope this helps. Good luck on the MCAT!

Someone correct me if I'm wrong.
 
  • #3
PhanthomJay
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You can't use your kinematic equation for constant acceleration when the force, and hence the acceleration, is not constant for the entire 10 m displacement. The problem is more easily solved using the work-energy theorem. Are you familiar with it?
 
  • #4
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Yes,

W total = ΔKE

so would it end up being along the lines of this?


W total = ΔKE

Fdcosθ= 1/2mv2f - 1/2mv2i

(10)F= 1/2(30)(2)2

F= 6 N?
 
  • #5
I don't believe force has anything to do with this problem?

It only asks for work, I got an answer of 60.0 J.
 
  • #6
PhanthomJay
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Yes,

W total = ΔKE

so would it end up being along the lines of this?


W total = ΔKE

Fdcosθ= 1/2mv2f - 1/2mv2i

(10)F= 1/2(30)(2)2

F= 6 N?
You are incorrectly assuming that W = Fd cosθ. That is for constant force only. Go back to your original attempt where you said that
I know how to solve the problem, I realize Figure 1: Force vs. Distance graph will give me the amount of work applied if I add the area under the graph.
Since the area under the graph is 1/2(30)(2)2 = 60 J, that's the work done by Alice, as whiskeySierra has also noted. Now apparently there is a part b to this problem that asks you to find the peak value of the Force. You should be able to find Fmax knowing that the area under the 'curve' is 60 J.
 

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