# Sled pulled up incline at an angle with friction

1. Oct 12, 2011

### Strawer

1. The problem statement, all variables and given/known data

A child pulls a 3.6 kg sled at 25 degrees to a slope that is 15 degrees to the horizontal, as in the figure. The sled moves at a constant velocity when the tension is 16 N. What is the acceleration of the sled if the rope is released?

2. Relevant equations

F=ma, f=$\mu$N

3. The attempt at a solution

Constant velocity means that friction + the component from Fg shall equal FTx (tension component parallell to plane) that is FTx = f + Fx

Where FTx=cos(25)*FT=14.5 N and Fx=3.6*9.8=9.13 N

Now I want $\mu$ so that I can calculate f. And we know that N=cos(15)*mg+T*sin(25)=40.84 N

If we go back to the start, FTx=$\mu$N+Fx so $\mu$= $\frac{ (FTx-Fx) }{N}$= $\frac{14.5-9.13}{40.84}$ = 0.131

When we release the sled N will reduce to N=mg*cos(15)=34.08 and so f=0.131*34.08=4.46 N

Fres=Fx-f=9.13-4.46=4.66 N

And the acceleration is a=$\frac{F}{m}$=$\frac{4.66}{3.6}$=1.3 m/s^2

But apparently this is not correct. Could someone please tell me what I've missed, this question has been very frustrating for me :(

Also, negative signs tend to mess upp my Latex, I apologize.

2. Oct 12, 2011

### Strawer

Here's the figure btw.

#### Attached Files:

• ###### Incline.png
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3. Oct 18, 2011

### Strawer

Anyone?
According to the answer sheet the correct answer should be 0.67 m/s^2 but i'm starting to wonder whether that's true or not?