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:(adsbygoogle = window.adsbygoogle || []).push({});

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 ?

Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Gravitational Potential Energy and Springs

**Physics Forums | Science Articles, Homework Help, Discussion**