You are an industrial engineer with a shipping company As part d the package-han
ID: 1600834 • Letter: Y
Question
You are an industrial engineer with a shipping company As part d the package-handling system, a small box with mass 1 30 kg is placed against a light spring that is compressed 0 280 m The spring has force constant 44 0 N/m. The spring and box are released from rest and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is mu_k = 0.300 When the box has traveled 0 280 in and the spring has reached its equilibrium length the box loses contact with the spring, What is the speed of the box at the instant when it leaves the spring? Express your answer with the appropriate units. What is the maximum speed of the box during its motion? Express your answer with the appropriate units.Explanation / Answer
A) Apply conservation of energy
net workdone = gain in kinetic energy
(1/2)*k*x^2 + fk.x = (1/2)*m*v^2
(1/2)*k*x^2 + mue_k*N*x*cos(180) = (1/2)*m*v^2
(1/2)*k*x^2 - mue_k*m*g*x = (1/2)*m*v^2
(1/2)*44*0.28^2 - 0.3*1.3*9.8*0.28 = (1/2)*1.3*v^2
==> v = 1.00 m/s
B)
the block gains maximum velocity
when spring force is equal to kinetic friction.
F_spring = Fk
k*x' = mue_k*m*g
x' = mue_k*m*g/k
= 0.3*1.3*9.8/44
= 0.08686 m
now again,
net workdone = gain in kinetic energy
(1/2)*k*(x^2 - x'^2) + fk.(x-x') = (1/2)*m*vmax^2
(1/2)*k*(x^2 - x'^2) + mue_k*N*(x- x')*cos(180) = (1/2)*m*vmax^2
(1/2)*k*(x^2 - x'^2) - mue_k*m*g*(x - x') = (1/2)*m*vmax^2
(1/2)*44*(0.28^2 - 0.08686^2) - 0.3*1.3*9.8*(0.28 - 0.08686) = (1/2)*1.3*v^2
==> v = 1.12 m/s
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