# MCAT Physics Question: Mechanics III

## 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

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

Related Introductory Physics Homework Help News on Phys.org
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$_{f}$-KE$_{i}$ = $\frac{1}{2}$mv$_{f}$$^{2}$-$\frac{1}{2}$mv$_{i}$$^{2}$

Hope this helps. Good luck on the MCAT!

Someone correct me if I'm wrong.

PhanthomJay
Homework Helper
Gold Member
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?

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?

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.

PhanthomJay
Homework Helper
Gold Member
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.