How far does the block slide? (work, spring, incline)

[magnesium12]

Homework Statement

The system is released from rest with no slack in the cable and with the spring stretched 225 mm. Determine the distance s traveled by the 3.2-kg cart before it comes to rest (a) if m approaches zero and (b) if m = 2.5 kg. Assume no mechanical interference and no friction. The distance s is positive if up the incline, negative if down.
Diagram: https://imgur.com/hniHevB

Homework Equations

W = Fd
W = ΔU + ΔT

The Attempt at a Solution

a)
I drew an FBD of the 3.2kg block, ignoring the other mass entirely. I had a F_s (force of the spring) pointing up the incline, F_W (weight of the block) pointing straight down, and N (normal force) pointing upwards normal to the incline.

I defined up the incline to be the positive x-axis. Normal to the incline is the y-axis.

I tried defining the potential energy states (U) and kinetic energy states (T) as it is released and as the block is released (state 1) and after it slides down the incline and comes to a rest (state 2).

U₁ = F_s1 = kx₁ = (160N/m)(0.225m) = 36N
U₂ = F_s2 = kx₂ = (160N/m)(s + 0.225m) = (160s)N + 36N
ΔU = U₂ - U₁ = 160s + 36 - 36 = 160s

T₁ = 0.5mv² = 0 (because v=0 right as it is released)
T₂ = 0.5mv² = 0 (because v=0 when it comes to a rest)
ΔT = 0

W = Fd = F_s(s) + F_ws = ks^s2 + mgs = 160s² + 3.2kg(9.81)s
= 160s² + 31.392s

W = ΔU = Fd
160s = 160s² + 31.392s
Solve for s: 0, 0.80379m (Incorrect Answer)

Can someone point out where I'm going wrong? Is it my interpretation of what s is? Or maybe how I'm calculating the potential and kinetic energies of each state?

b)
I drew another FBD which has the same components as in part a, but with an added uphill tension caused by the force of gravity acting on the second mass (called F_T).

Calculating F_T from the second mass:
2T + F_W,2 = 0
T = F_W,2 /2 = 2.5kg(9.81m/s/s)/2 = 12.263I tried defining the potential energy states (U) and kinetic energy states (T) as it is released and as the block is released (state 1) and after it slides down the incline and comes to a rest (state 2).

U₁ = F_s1 + F_T= kx₁ = (160N/m)(0.225m) = 36N + 12.263N = 48.263N
U₂ = F_s2 + F_T = kx₂ = (160N/m)(s + 0.225m) + 12.263N = (160s)N + 48.263N
ΔU = U₂ - U₁ = 160s + 48.263 - 48.263 = 160s

T₁ = 0.5mv² = 0 (because v=0 right as it is released)
T₂ = 0.5mv² = 0 (because v=0 when it comes to a rest)
ΔT = 0

W = Fd = F_s(s) + F_ws + F_Ts= ks^s2 + mgs + 12.263 = 160s² + 3.2kg(9.81)s + 12.263
= 160s² + 31.392s + 12.263

W = ΔU = Fd
160s = 160s² + 31.392s +12.263
Solve for s: 0.11056 OR 0.69323m (Incorrect Answers)

I assume I'm doing the same thing wrong here as in part a, but I can't figure out what it is.

Thank you for any help you can give!

 

[haruspex]

What is the elastic PE stored in a spring of constant k and extended or compressed distance x?

 

[Miles123K]

Is there a mass associated with the pulley? The system is frictionless right? Maybe first define the coordinates of the pulleys and draw free diagram for each one?

 

[magnesium12]

What is the elastic PE stored in a spring of constant k and extended or compressed distance x?

Thank you for your help!

Okay, so I figured out the PE of a spring is kx²/2. So for the first case (ignoring the second mass completely), I've now got this:

Potential energy of the block
U₁ = kx²/2 = 160(0.225)²/2 = 4.05
U₂ = ks² = 160(s)²/2 = 80s²
ΔU = U₂ - U₁ = 80s² - 4.05

Work done on the block by gravity:
W_g = -mg_xs = -mgsin(26)s (assuming up the incline is the x-axis)

Work done on the block by the spring:
W_s = F_ss = (ks)s = 160s²

ΔU = ΣW
80s² - 4.05 = 160s² - mgsin(26)s
0 = 80s² - 3.2(9.81)sin(26)s + 4.05 <---------------------- gives no solutions.

Is there some other component doing work on the system that I am missing? Or do I need to take into consideration the kinetic energy too? If so, do I need to assume there is a velocity for the block right as the system is released?

 

[haruspex]

Work done on the block by the spring:
Ws = Fss = (ks)s = 160s2

You are making the same mistake again. The force the spring exerts is not constant across the range s.