Physics: Springs: A block of mass 1.4 kg is attached to a horizontal spring that has a force constant 1100 N/m as shown in the figure...?

A block of mass 1.4 kg is attached to a horizontal spring that has a force constant 1100 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest.

https://imgur.com/kkAqguN

(a) A constant friction force of 4.2 N retards the block's motion from the moment it is released. How much is the spring compressed when the speed of the block is a maximum.

____cm

(b) What is the maximum speed?

____cm/s

2 Answers

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  • NCS
    Lv 7
    1 month ago
    Best Answer

    initial energy in the spring is

    Ei = ½kx² = ½ * 1100N/m * (0.020m)² = 0.22 J

    At lesser compression, that initial energy has been converted into friction work, KE, and some remaining spring energy:

    E = 4.2N(0.020m - x) + ½mv² + ½kx²

    and that total must equal Ei.

    So

    4.2N(0.020m - x) + ½mv² + ½kx² = 0.22 J

    means that (dropping units for ease)

    ½mv² = 0.22 - ½kx² - 4.2*(0.020 - x)

    differentiate w/r/t x

    ½m * 2v * dv/dx = -kx + 4.2

    we want the point where dv/dx = 0, which is where

    x = 4.2 / k = 4.2 / 1100 = 0.0038 m = 0.38 cm

    ½mv² = 0.22 - ½*1100*0.0038² - 4.2*(0.020 - 0.0038)

    ½ * 1.4kg * v² = 0.144 J ← just over this value, actually

    v² = 0.206 m²/s²

    v = 0.45 m/s ◄

    it always makes me nervous when I do dv/dx and not dv/dt. Let's check:

    at x = 0.0035 m,

    ½mv² = 0.22 - ½*1100*0.0035² - 4.2*(0.020 - 0.0035) = just under 0.144 J

    and at x = 0.0040 m,

    ½mv² = 0.22 - ½*1100*0.0040² - 4.2*(0.020 - 0.0040) = 0.144 J

    and so it seems to me that we have in fact found the max velocity at x = 0.0038 m.

    If you find this helpful, please award Best Answer. You get points too!

  • Anonymous
    4 weeks ago

    not a question. reported!

    • Anonymous4 weeks agoReport

      Problems are not questions but they still need answers.

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