Question

If an object attached to a hanging unstretched ideal and massless spring is slowly lowered to its equilibrium position, a distance of 6.4 cm below the starting point, how far would it stretch the spring at maximum elongation if instead, the object was dropped from rest?

Option 1: 3.2 cm
Option 2: 6.4 cm
Option 3: 9.6 cm
Option 4: 12.8 cm

Answer 1

When an object is dropped from rest and it is attached to a hanging spring, it will stretch the spring at maximum elongation by 8 cm. no Option is correct.

Step-by-step explanation:

To determine the maximum elongation of the spring when the object is dropped from rest, we can use the principle of conservation of energy. The initial potential energy of the object is converted into the maximum potential energy of the spring. We can equate these two energies:

mgh = (1/2)kx^2

where m is the mass of the object, g is the acceleration due to gravity, h is the height the object is dropped from, k is the force constant of the spring, and x is the maximum elongation of the spring.

By solving for x, we can find the maximum elongation of the spring. Substituting the given values into the equation, we have:

(1/2)kx^2 = mg(2h)

x^2 = 4gh

x = √(4gh)

Plugging in the values for g and h (the distance the object is dropped), we find:

x = √(4 * 9.8 m/s^2 * 0.064 m) = 0.8 m = 8 cm

Therefore, the object dropped from rest would stretch the spring at maximum elongation by 8 cm.