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Gravitational Potential Energy and Springs

  1. Nov 20, 2006 #1
    A 520 kg object is released from rest at an altitude of 800 km above the north pole of the earth. Ignore atmospheric friction. The speed of the object as it strikes the earth’s surface, in km/s, is closest to:

    a.3.7----b. 2.7---c.1.9-----d. 4.0------e. 2.3

    Conservation of energy, where:

    -G*(m_E*m)/(r + a_1) = 0.5*m*v^2 + -G*(m_E*m)/(r )

    r = 6.38*10^6 m
    a1 = 800,000 m

    Isolating v:
    -G*(m_E*m)/(r + a_1) + G*(m_E*m)/(r ) = 0.5*m*v^2

    2*G*(m_e*m)*[(-1/(r_e + a1)) + (1/r_e)] = m*v^2

    v = sqrt[2*G*(m_e)*[(-1/(r_e + a1)) + (1/r_e)]

    Plugging in the values:
    v = 3729.4 m/s??? (before converting to km)

    An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 5.1 cm. The force constant of the spring is 1900 N/m. The bullet’s initial velocity is closest to:
    a.600 m/s-----b. 580 m/s-----c. 530 m/s------d. 560 m/s-----e. 620 m/s

    conservation of energy:

    0.5*k*x^2 = 0.5*(m_bullet + m_block)*v^2 ?

    v_final = sqrt[kx^2/(m_block + m_bullet)], where m_bullet = .008 kg, x = 0.05 m, m_block = 4.0 kg?

    V_final = 1.11 m/s???

    V_initial = (m_block +m_bullet)*v_f/(m_bullet) = 556.3 m/s ?

  2. jcsd
  3. Nov 21, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I didn't check your calculations, but your solutions to both problems look good to me.
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