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A rubber band slingshot is drawn back 0.15 m with a force of 28N and shoots 28 g stone the initial speed of the stone if the rubber band is released is

Respuesta :

Answer:

Initial speed of stone = 12.25 m/s

Explanation:

Given:

Change in distance = 0.15 m

Force applied = 28 N

Find:

Initial speed of stone

Computation:

Using conservation of energy

(1/2)fs² = (1/2)(mv²)

fs² = (mv²)

(28)(0.15)² = (0.028)(v²)

v² = (28)(0.15)²/(0.028)

v = 12.25

Initial speed of stone = 12.25 m/s

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