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Sonja [21]
2 years ago
7

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

we release the spring, when the spring returns to its normal length, how many joules is the kinetic energy of the lighter body?
Physics
1 answer:
cluponka [151]2 years ago
7 0

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

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Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
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Dado lo siguiente:

Altura desde la cual se cae el cuerpo = 120 m

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A.) La velocidad que toma:

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B) Cuánto ha disminuido.

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If two cars with the same mass( cars a and b) are moving but one (car b) is moving at twice the speed of car a ?
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