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25 July, 00:21

An object whose weight is 100lbf (pound force) experiences a decrease i kinetic energy of 500ft-lb, and an increase in potential energy of 1500ft-lbf. The initial velocity and elevation of the object, each relative to the surface of the Earth are 40ft/s and 30ft.

(a). Find final velocity in ft/s. (b) find final elevation.

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Answers (2)
  1. 25 July, 00:43
    0
    A. 43.84 ft/s.

    B. 45 ft.

    Explanation:

    A.

    Given:

    Weight, W = 100 lbf

    mass, M = 100/32.174

    = 3.108 lb

    kinetic energy, ΔKE = - 500 ft. lbf

    Potential energy ΔPE = 1500 ft. lbf

    initial velocity, u = 40 ft/s

    initial height, h1 = 30 ft/s

    acceleration, g = 32.174 ft/s^2

    ΔKE = 1/2 * M * (v^2 - u^2)

    -500 = 1/2 * 3.108 * (v^2 - 40^2)

    v = 43.84 ft/s.

    B.

    ΔPE = M * g * (h2 - h1)

    1500 = 3.108 * 32.174 * (h2 - 30)

    h2 = 45 ft.
  2. 25 July, 00:48
    0
    a) 35.75 ft/s

    b) 45 ft

    Explanation:

    Given

    Weight W = 100 lbf

    mass (m) = 100*32.174/32.2=99.92 lb

    decrease in kinetic energy ΔKE = - 500 ft. lbf

    increase in kinetic energy ΔPE = 1500 ft. lbf

    initial velocity V_1 = 40 ft/s

    initial height h_1 = 30 ft/s

    The gravitational acceleration g = 32.2 ft/s2 Required

    (a) Final velocity V_2 (a) Final elevation h_2

    Solution

    Change in kinetic energy is defined by

    ΔKE =.5*m * (V_2 ^2-V_1^2)

    Change in potential energy is defined by

    ΔKE = W * (h_2 - h_1)

    Then,

    -500=.5*99.92*1/32.174 * (V_2 ^2-40^2)

    V_2=35.75 ft/s

    1500 = 100 x (h_2 - 30)

    h_2 = 45 ft
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