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# Physics homework

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Aghhh....i think I've finally crashed...i can't seem to figure a problem out...actually two please anyone help?

1. A 5.0 kg block is sent sliding up a plane inclined at 37 degrees while horizontal force F of magnitude 50N acts on it. The coefficient of kinetic friction is .30. a) magnitude of the block's acceleration

Ok i know the direction of the acceleration is down the incline...but i'm really stuck on the equation to find the acceleration....

2. A 1.34kg ball is connected by means of two massless strings, each of length L=1.7m to a vertical, rotating rod. The strings are tied to the rod with separation of d=1.70 m and are taut. The tension in the upper string is 35N. a) the tension in the lower string, :) the magnitude of the net string force on the ball c) speed of the ball

I can get c, if i can get a)...

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just to let everyone know that i'm not shamming....lol this is what i have for the first problem:

@=theta..

U=coefficient of kinetic friction

in the x-dir

FCos@-WSin@+UN=ma

in the y-dir

N-FSin@+WCos@=ma

N=FSin@+WCos@

So the equation by substituting N would be:

FCos@-WSin@+U(FSin@+WCos@)=ma

and acceleration according to this would just be all that divided by m....but something must be off bcuz that's not the correct answer...:)

btw, the acceleration should be 2.1 m/s*s

please someone tell me what is wrong with my equation? lol i'm desperate to know what i don't understand

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The force of friction (UN) is in the negative x direction so you

need to change the x-dir equation to

FCos@-WSin@-UN=ma

There's no acceleration in the y direction so the equation is

N-FSin@-Wcos@=0

N=FSin@+WCos@

So you get

FCos@-WSin@-U(FSin@+WCos@)=ma

a=1/5.0*(50Cos37-49Sin37-.3(50Sin37+49cos37))

a=-2.1

2.1 m/s^2 going down the incline

I haven't looked at the 2nd question yet, just a sec.

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Ok, the spinning string and ball system forms an equilateral triangle with the vertical rotating rod. If you set up the force equations like before, let T1=tension on the top string=35N and T2=tension on the bottom string, then

x-dir: F=T1cos30+T2cos30=ma=m(v^2/r) (r=LCos30=1.7cos30=1.5m)

y-dir: T1sin30-T2sin30-W=0 (W=mg=1.43kg*9.8m/s^2)=14N)

T2=(T1sin30-W)/sin30=7N

F=36N net string force on the ball

v=sqr(F*r/m)=6.1m/s speed of the ball

Have fun :)

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haha yeah i eventually figured out that coefficient of friction was negative. Thanks for the second problem! I didn't know where to start the problem from. That helps alot

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