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magnesium12
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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 Fs (force of the spring) pointing up the incline, FW (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 = Fs1 = kx1 = (160N/m)(0.225m) = 36N
U2 = Fs2 = 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 = Fs(s) + Fws = kss2 + 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 FT).
Calculating FT from the second mass:
2T + FW,2 = 0
T = FW,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 = Fs1 + FT= kx1 = (160N/m)(0.225m) = 36N + 12.263N = 48.263N
U2 = Fs2 + FT = 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 = Fs(s) + Fws + FTs= kss2 + 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!