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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?

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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(A) The average velocity in kilometres/hour

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(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

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$128.74 \times \frac{5}{18}$

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kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

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Substituting the known values,

$s= ut+\frac{1}{2}at^2$

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v=u +at

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