Item 1 The 5-lb box slides on the surface for which Mu k = 0.3. The box has a ve

Question

Item 1 The 5-lb box slides on the surface for which Mu k = 0.3. The box has a velocity v = 15 ft/s when it is 2 ft from the plate. Part A If the box strikes the smooth plate, which has a weight of 21 lb and is held in position by an unstretched spring of stiffness k = 480 lb/ft, determine the maximum compression imparted to the spring. Take e = 0.8 between the box and the plate. Assume that the plate slides smoothly. Express your answer using three significant figures and include the appropriate units. X = 1.09 ft

Explanation / Answer

First, we will calculate velocity of box before collision,

Work done by friction = dot product (f *x) = - 0.3* 5 * 2 = 3 lbf     ,,,,,,f = mu * m*g

Change in K.E. of block = - 3 lbf

Kinetic energy at beginning = 0.5 * m * v^2 = 0.5 * (5/32.2) * 15^2 = 17.47 lbf

Kinetic energy before collion = 17.47 -3 = 14.47 lbf

Velocity before collision = 0.5 * (5/32.2) * (v2)^2 = 14.47

v2 = 13.65 ft/s

Velocity just after collision of box, v3

Velocity before and after collision of plate be v' and v"

Using Conservation of momentum

m1 * v2 + m2 * v' = m1 * v3 + m2*v"

5*13.65 + 0 = 5 * v3 + 21 *v" ....1)

Also, (v" - v3)/v2 = 0.8

v" - v3 = 10.92.....2)

Using 1 & 2

68.25 = 5 ( v" - 10.92) + 21v"

v" = 4.725 ft/s

Using energy conservation,

0.5*m * v" ^2 = 0.5 * k*x^2

(21/32.2) * (4.725)^2 = 480 * x^2

x = 0.174 ft

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