Gravitational Potential Energy and Springs

[Soaring Crane]

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 ?

Thanks.

 

[Doc Al]

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

 

[kyle8921]

I would like to clarify that the answer given for the initial velocity of the bullet (556.3 m/s) is incorrect. The correct answer should be 556.2 m/s, as calculated using the conservation of momentum equation: m_bullet*v_initial = (m_block + m_bullet)*v_final. This calculation takes into account the fact that the bullet remains lodged in the block and the total mass of the system changes. Additionally, the final velocity should be rounded to 1.11 m/s, not 1.10 m/s. It is important to be precise in our calculations and provide accurate answers.