Aphysics student hangs various masses from a spring and records the resulting distance the spring has stretched. the graph shows the data, which plots the mass hung from the spring as a function of the distance the mass caused the spring to stretch beyond its original length. by how much will the spring have stretched if you hang a 2.35 kg mass from it, assuming that it continues to obey hooke’s law? 120 cm 94 cm 76 cm 50 cm

94 cm

Explanation:

The answer is not among the given options, however it can be approximated to 94 cm.

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force applied to it, as long as the spring is not permanently deformed:  

(1)

Where:  

is the elastic constant of the spring

is the length of the spring without applying force.  

is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.  In other words, the force applied is the weigth :

(2)

So, equation (1) can be rewritten as:

(3)

Isolating :

(4)  From here we can deduce is the slope of the line of the graph shown, as follows:

Where:

is the acceleration due gravity

(5) This is the spring constant

Now, with the mass we can calculate the weight:

(6)

So, knowing and we can calculate from (3):

(7)

(8)

(9)

The value that is nearest to 88.13 cm is 94 cm

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