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18 September, 02:21

The weight of the pendulum of an impact tester is 451b, and the length of the 27 pendulum arm is 32in. If the arm is horizontal before striking the specimen, what is the potential energy of the tester? What is the striking velocity?

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Answers (2)
  1. 18 September, 02:37
    0
    U = 120 lb*ft = 162.698 J

    v = 13.105 ft/s

    Explanation:

    Given

    W = 45 lb

    L = 32 in

    U = ?

    v = ?

    We apply the equation

    U = W*h

    where h = L = 32 in * (1 ft/12 in) = 2.66 ft

    ⇒ U = 45 lb * 2.66 ft = 120 lb*ft

    then, we apply

    U_in = K_in

    ⇒ W*h = m*g*h = 0.5*m*v²

    ⇒ v = √ (2*g*h)

    ⇒ v = √ (2*32.2 ft/s²*2.66 ft) = 13.105 ft/s
  2. 18 September, 02:46
    0
    4 m/s.

    Explanation:

    Given:

    W = 45 lb

    Converting lb to kg,

    45 lb = 20.41 kg

    L = 32 in

    Converting in to m,

    32 in = 81.3 cm

    = 0.813 meters

    A.

    PE = mgh

    = 20.41 * 9.81 * 0.813

    = 162.78 J

    B.

    Change in PE = Change in KE

    m*g*h = 0.5*m*v²

    v = √ (2*g*h)

    9.8 * 0.813 = 0.5 * v²

    v = √ (2 * 9.8 * 0.813)

    = 4 m/s
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