Gravitational Potential Energy and Springs

In summary, for the first problem, the speed of the object as it strikes the earth's surface, after being released from rest at an altitude of 800 km above the north pole of the earth and ignoring atmospheric friction, is closest to 3729.4 m/s (or 3.7 km/s). For the second problem, the initial velocity of an 8 g bullet shot into a 4.0 kg block, which then moves into a spring and compresses it by 5.1 cm, is closest to 556.3 m/s.
  • #1
Soaring Crane
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0
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.
 
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  • #2
I didn't check your calculations, but your solutions to both problems look good to me.
 
  • #3


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.
 

Related to Gravitational Potential Energy and Springs

What is gravitational potential energy?

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is the energy that is stored when an object is lifted to a certain height above the ground.

How is gravitational potential energy calculated?

The gravitational potential energy of an object is calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.

What is the relationship between gravitational potential energy and springs?

Gravitational potential energy and springs are related through the concept of potential energy. Just as objects have gravitational potential energy due to their position in a gravitational field, objects attached to springs have potential energy due to their position relative to the spring's equilibrium point.

How does the spring constant affect gravitational potential energy?

The spring constant, which represents the stiffness of a spring, affects gravitational potential energy by determining how much potential energy is stored in the spring when it is stretched or compressed. A higher spring constant results in a larger potential energy for a given displacement.

Can gravitational potential energy be converted into other forms of energy?

Yes, gravitational potential energy can be converted into other forms of energy, such as kinetic energy. When an object falls, its potential energy decreases and is converted into kinetic energy. It can also be converted into thermal energy through friction or mechanical energy through the use of machines.

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