# Work and Energy

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1. Oct 29, 2016

### DCholic

1. The problem statement, all variables and given/known data
Masses A and B, each having a mass of 32.2 slugs, are constrained to move in frictionless slots. They are connected by a rigid, massless rod of length L = 2 ft. Mass B is connected to two massless linear springs, each having a spring constant k = 60 lbF / ft. The springs are unstretched when the connecting rod to masses A and B is vertical.

The accompanying schematic:

If B is given a slight nudge, what are the velocities of B and A when A descends a distance of D = 3 inches?

2. Relevant equations
W1->2 = ΔKE + ΔPEG + ΔPEE

3. The attempt at a solution
Masses: mA = mB = m
Take upward to be positive y and right as positive x.

At position 1:
Velocity of A: vA1 = 0
Velocity of B: vB1 = 0
Height of A: hA1 = L
Height of B: hB1 = 0
Displacement of left spring: xL1 = 0
Displacement of right spring: xR1 = 0

Finding the displacement of each spring:
(L - D)2 + x2= L2
x = 11.62 in. = 0.9682 ft -> symmetric for both springs

At position 2:
Velocity of A: vA2 = ?
Velocity of B: vB2 = ?
Height of A: hA2 = L - D
Height of B: hB2 = 0
Displacement of left spring: xL2 = x
Displacement of right spring: xR2 = x

Work:
W1->2 = 0 because there are no non-conservative forces acting on the system

Change in kinetic energy:
ΔKE = (1/2) * mA * (vA22 - vA12) + (1/2) * mB * (vB22 - vB12)
ΔKE = (1/2) * m * (vA22 + vB22)

Change in gravitational potential energy: g = 32.174 ft/s2
ΔPEG = mA * g * (hA2 - hA1) + mB * g * (hB2 - hB1)
ΔPEG = m * g * ( - D) = -259.001 slug⋅ft2/s2 = -259.001 ft⋅lbF

Change in elastic potential energy:
ΔPEE = (1/2) * k * (xL22 - xL12) + (1/2) * k * (xR22 - xR12)
ΔPEE = (1/2) * k * ( 2 * x2)
ΔPEE = k * x2 = 56.245 ft⋅lbF

Relating vA2 and vB2:
Take the origin at B's initial position.
x ≡ horizontal distance from origin to B
y ≡ vertical distance from origin to A

x2 + y2 = L2 = constant
2 * x * (dx/dt) + 2 * y * (dy/dt) = 0
2 * x * vB + 2 * y * vA = 0
x * vB + y * vA = 0

At the instant when A has moved down 3 inches:
vB2 = [ - (L - D) / x] * vA2 = -1.807 * vA2

Combine all equations:
W1->2 = ΔKE + ΔPEG + ΔPEE
0 = (1/2) * m * [vA22 + ( -1.807 * vA2)2] - 259.001 + 56.245
vA2 = 1.718 ft/s down
vB2 = 3.105 ft/s right

This was a midterm question. Physics enthusiasts, please give feedback my thought process. Much appreciated.

2. Oct 29, 2016

### TSny

Welcome to PF!

Nicely organized. Your work looks correct to me.