Question

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched by a length of 4.3 cm. How long is the spring when a 3.3 kg mass is suspended from it in cm?

Answer 1

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$