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Jul 1, 2008

Problem 3/134

Is the work done on the 15 kg block + 15 kg sleeve equal to the work done on the 50 kg block since they are in the same system? Is there even a connection or do you need to find 1/2mv^2=Fs to find s for the 50 kg block?
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2 comments:

ckearney said...

I have been trying to describe this system for the last few hours as well. The simple setup which will allow me to solve is not yet apparent.

What I have done so far involves recognizing that the 50-kg block will experience a change in acceleration when the 15-kg collar is removed (two separate accelerations total). Therefore the distance "s" is composed of s1 and s2. I have written Newtonian (F=ma), Work-Energy (1 to 2, 2 to 3), and Kinematics (constant length cord) equations to relate the cylinder+collar and block. While I have solved for T (tension) at the first and second states (T1 and T2), the acceleration of the 50-kg block I get using T1 is negative (= down the incline) and using T2 the acceleration is positive.

It doesn't quite make physical sense for the block to move down the incline, and then, with even less mass producing smaller Tension, go back up the slope. Here are the details:

T1 = 109.32 N
a_x' = -3.084 [m/s^2]
T2 = 382.35 N
a_x'' = +7.84 [m/s^2]

By writing two Work-Energy equations for the 50-kg block, and trudging through some ridiculous looking algebra, I was able to obtain the following values:

s1 = -1.768 [m]
s2 = -1.857 [m]

s = s1 + s2
s = -3.62 [m]

Although I am worried about being in the dark as to the simple solution, I have a answer that doesn't look too ridiculous.

Good luck.

July 2, 2008 at 12:41 AM
Thoesen said...

So in other words its so ridiculous that we shouldn't worry about it?

July 2, 2008 at 2:33 PM

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