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When a 4.20-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.60 cm. (a) If the 4.20-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it

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muhammadjonedkhattak

Answer:

Y = 0.009 m = 0.9 cm

Explanation:

Mass m = 4.20 Kg,

Stretched spring Distance = x= 2.60 cm = 0.026 m

Weight = F= m g = 4.20 Kg × 9.81 m/s² = 41.202 N

Hook's Law F= K x

K = F/x = 41.202 N / 0.026  m

K = 1584.69 N/m

To Find the compressed spring distance Y=? when 1.5 kg block still hung on it So F = mg = 1.5 x 9.81 = 14.715 N

Y = F/k =  14.715 N / 1584.69 N/m

Y = 0.009 m = 0.9 cm

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