Determine the velocity of the Slider

[Northbysouth]

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 + U_1-2 = 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)v²_b - 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.

 

[Simon Bridge]

Please explain your reasoning behind the numbers you wrote down.

 

[haruspex]

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))

 

[Muhammad2548]

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

 

[Steve4Physics]

Hi @Muhammad2548. Welcome to PF.

You may want to note that the thread is over 11 years old!

 

[Muhammad2548]

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.