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

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.