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Sidana [21]
3 years ago
14

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

g friction of the horizontal surface that the block is sliding on, how fast is the block moving once it crosses back by the equilibrium position after being released
Physics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as mgh=0.5mv^{2} since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then v=\sqrt {2gh} and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s

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