When a 4.48 kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.68 cm.

How much work must an external agent do to stretch the spring 3.96 cm from its unstretched position?

from the mass (remember mg is weight), find the spring constant.

Work=1/2 kx^2

128.45

To calculate the work done in stretching the spring, we need to use Hooke's law and the formula for work.

Hooke's law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as: F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

The formula for work is: W = (1/2)kx^2, where W is the work done, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, we are given the mass of 4.48 kg hanging vertically on the spring, resulting in a displacement of 2.68 cm. We can use this information to calculate the spring constant (k) using the equation F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

First, convert the displacement from centimeters to meters: x = 2.68 cm = 0.0268 m.

Next, calculate the force exerted by the spring using the equation F = mg: F = (4.48 kg)(9.8 m/s^2) = 43.904 N.

Now, we can calculate the spring constant (k) using Hooke's law: k = F/x = 43.904 N / 0.0268 m = 1637.3134 N/m.

Finally, we can calculate the work done to stretch the spring 3.96 cm from its unstretched position using the formula for work: W = (1/2)kx^2.

First, convert the displacement from centimeters to meters: x = 3.96 cm = 0.0396 m.

Now, calculate the work done: W = (1/2)(1637.3134 N/m)(0.0396 m)^2 = 0.49387 J.

Therefore, an external agent must do approximately 0.49387 Joules of work to stretch the spring 3.96 cm from its unstretched position.