Aspiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . when an object of mass of 20g is placed in pan ,its length becomes 21.80cm ,when another object of mass 30g is placed in the pan, the length, becomes 11.05cm. calculate the mass of the scale pan.

Explanation:

There seems to be a typo in the problem statement.  It says the spring stretches to a shorter length after more mass is added.  Please check the problem statement.  I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm.  The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables.  If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

i think the answer is a, hope this is right, that would be my best guess.

where are the vectors ?