A block is released from rest, determine its velocity

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Alexanddros81
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Homework Statement


14.10 Solve Prob. 12.47 by the work-energy method

12.47 When the 1.8-kg block is in the position shown, the attached spring is
undeformed. If the block is released from rest in this position, determine its velocity
when it hits the floor

Diagram of a block hanging from a vertical spring with stiffness k = 80 N/m, the block labeled 1.8 kg, a 0.3 m vertical gap to the floor marked below it, and a coordinate x pointing right.


Homework Equations

The Attempt at a Solution


Here is my solution to both 12.47 and 14.10.
I get in 14.10 the correct answer but I am confused with 12.47. Is it correct?
Handwritten work-energy solution to problem 14.10: U(A→B) = T_B − T_A, giving U(A→B) = ½mv_B² − 0; work by weight and spring; v_B² = 2x(mg − ½Kx)/m = 0.6(17.658 − 12)/1.8 = 1.886, so v = 1.373 m/s.
Handwritten solution to problem 12.47: a free-body diagram of a block with spring force Kx upward, weight W and a kinematic diagram labeled 0.3 m, followed by the equations mg - Kx = ma and v^2 = 2(0.3)(1.8·9.81 - 80·0.3)/1.8, then v =
 
Attachments
  • Handwritten solution to problem 12.47: a free-body diagram of a block with spring force Kx upward, weight W and a kinematic diagram labeled 0.3 m, followed by the equations mg - Kx = ma and v^2 = 2(0.3)(1.8·9.81 - 80·0.3)/1.8, then v =
    Pytels_Dynamics108.jpg
    19.3 KB · Views: 696
  • Handwritten work-energy solution to problem 14.10: U(A→B) = T_B − T_A, giving U(A→B) = ½mv_B² − 0; work by weight and spring; v_B² = 2x(mg − ½Kx)/m = 0.6(17.658 − 12)/1.8 = 1.886, so v = 1.373 m/s.
    Pytels_Dynamics109.jpg
    19.7 KB · Views: 609
  • Diagram of a block hanging from a vertical spring with stiffness k = 80 N/m, the block labeled 1.8 kg, a 0.3 m vertical gap to the floor marked below it, and a coordinate x pointing right.
    Fig P12_47.jpg
    4 KB · Views: 712

Discussion

When you wrote
Handwritten derivation: a = v dv/dx, so a dx = v dv, and the integral from 0 to Δx of a dx equals the integral from 0 to v of v dv.

you assumed ##\int a dx = a \Delta x##. Is this true if the acceleration is not constant?

What is the main topic of chapter 12?
 
Attachments
  • Handwritten derivation: a = v dv/dx, so a dx = v dv, and the integral from 0 to Δx of a dx equals the integral from 0 to v of v dv.
    upload_2017-12-18_13-7-12.png
    3.5 KB · Views: 566
Handwritten derivation: sum of forces in the downward direction gives mg - Kx = ma, then a = v dv/dx, leading to integration yielding v^2/2 = xg - Kx^2/2m + C, with C = 0 when x = 0 and v = 0, so v^2 = 2gx - Kx^2/m and v = sqrt(2gx - Kx^2/m) = 1.373 m/s.


Hi! After reading a sample problem of the book I came up with the above solution.
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
 
Attachments
  • Handwritten derivation: sum of forces in the downward direction gives mg - Kx = ma, then a = v dv/dx, leading to integration yielding v^2/2 = xg - Kx^2/2m + C, with C = 0 when x = 0 and v = 0, so v^2 = 2gx - Kx^2/m and v = sqrt(2gx - Kx^2/m) = 1.373 m/s.
    Pytels_Dynamics111.jpg
    18.4 KB · Views: 1,024
Alexanddros81 said:
View attachment 217021
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
Each side of the equation would have its own constant of integration. You can combine the two constants into one constant. Then, this constant is determined by the initial conditions as you have done.

Your work here is essentially the derivation of the work-energy principle for this particular problem.
 
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Ok thanks for that.
I need to revise my calculus.

Alexandros
 

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