Determine the velocity of the Slider

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The discussion revolves around calculating the velocity of a 0.6 lbf slider moving along a curved rod under a constant tension of 1.3 lb. The initial approach involved using energy conservation principles, but the user encountered confusion regarding the calculations and the interpretation of terms. Key corrections included recognizing the height as 10 inches and recalculating the work done by the force. The final calculations led to a velocity of approximately 13.64 ft/s. The thread serves as a resource for others facing similar physics problems.
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Homework Statement


The 0.6 lbf slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 1.3 lb tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B.


I have attached an image of the question

Homework Equations





The Attempt at a Solution



As far as I can tell this is an energy problem, hence:

T1 + V1 + U1-2 = T2 + V2

T1 and V1 are zero as there is no initial energy, spring forces or gravitational forces at A.

Hence, I get:



(13/12)sin(67.38) = 1/2*(.6/32.3)v2b - 0.6*(10/32.3)

v = 11.28

Not quite the answer I'm looking for and I'm a little unsure where I'm making my mistake.

Any help would be appreciated.
 

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Please explain your reasoning behind the numbers you wrote down.
 
Ignore everything left of the pulley for a moment. If you just watched the string coming over the pulley, how would you figure out the work done by the force?
(Also, I don't understand the division in this term: 0.6*(10/32.3))
 
First of all apparently the height equals 10 inches not feet.
So the horizontal distance = (24-6)/12 = 1.5 feet and the vertical distance = 5/6 feet
With that said let us solve it using the formula that work equals change in energy.

W = K.E.2 + U2 - (K.E.1 + U1)

As you said K.E.1 + U 1 equals 0 so let's figure out the other values

W = F*S*cosθ, F= 1.3, S= (1.52 + (5/6)2)1/2, θ = 0
So W ≈ 2.23

K.E.2 = 1/2 mv2, m = 0.6/32.3. So K.E. = 3/323 v2

U2 = mgh = 0.6*(5/6)= 0.5

So 2.23 = 3/323 v2 + 0.5
1.73 = 3/323 v2
v2 ≈ 186.26
v ≈ 13.64
 
Hi @Muhammad2548. Welcome to PF.

You may want to note that the thread is over 11 years old!
 
Steve4Physics said:
Hi @Muhammad2548. Welcome to PF.

You may want to note that the thread is over 11 years old!
Yeah I made it for anyone unable to answer it like I was.
 
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