When a 3.50-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.10 cm. (a) If the 3.50-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?

Explanation:

Hooke's law state that, provided the elastic limit of an elastic material is not exceeded, the extension e is directly proportional to the applied force.

given data

mass m= 3.5kg

extension  e= 2.10cm to m = 2.1/100=0.021

let us solve for the spring constant k

we know that

F=keHooke's law

F=ma1st law of motion

ma=ke

k=ma/e

Assume a=10m/s^2

k=3.5*10/0.021

k=35/0.021

k=1666.66N/m

(a) If the 3.50-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it

so mass is now= 1.5kg

we found k = 1666.66N/m

ma=ke

e=ma/k

e=1.5*10/1666.66

e=15/1666.66

e=0.009m

in cm =0.90cm

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