Answer:
7.5 cm
Explanation:
The force exerted by the 2.0 kg on the spring is equal to the weight of the mass, so:
[tex]F=mg = (2.0 kg )(9.8 m/s^2)=19.6 N[/tex]
And this force causes a stretching of:
[tex]\Delta x = 15 cm - 10 cm = 5 cm[/tex]
So we can use Hooke's law:
[tex]F=k \Delta x[/tex]
to find k, the spring constant:
[tex]k=\frac{F}{\Delta x}=\frac{19.6 N}{5 cm}=3.92 N/cm[/tex]
Now a new mass of m = 3.0 kg is attached to the spring; the force applied by this mass is
[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]
So we can use again Hooke's law to find the new stretching:
[tex]\Delta x = \frac{F}{k}=\frac{29.4 N}{3.92 N/cm}=7.5 cm[/tex]
