Question

A spring stretches 5.52 cm vertically when a 2.50 kg object is suspended from it. Find the distance (in cm) the spring stretches if you replace the 2.50 kg object with a 3.40 kg object. Assume that the spring obeys the Hook's law.

Answer 1

7.51 cm

Step-by-step explanation:

y1 = 5.52 cm

m1 = 2.5 kg

m2 = 3.4 kg

Let the spring stretches by the distance y2.

According to Hooke's law

F = - k y

So,
`(F_(1))/(F_(2))=(y_(1))/(y_(2))`

`(2.5 * 9.8)/(3.4 * 9.8)}=(5.52)/(y_(2))`

y2 = 7.51 cm

Thus, the spring is stretches by 7.51 cm.