Final answer:
A spring with a force constant of 44.5 N/m will be stretched by approximately 0.10 m when an object with a mass of 0.450 kg is hung motionless from the spring.
Step-by-step explanation:
In order to determine how much a spring will be stretched by an object with a given mass when hung motionless from the spring, we need to use Hooke's Law, which states that the force applied by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula is F = -kx, where F is the force applied by the spring, k is the force constant, and x is the displacement from equilibrium.
In this case, the force constant is given as 44.5 N/m and the object has a mass of 0.450 kg. Since the object is motionless, we know that the net force on the object will be zero, which means the force applied by the spring is equal to the gravitational force acting on the object.
Using Newton's second law, F = mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2), we can set up the following equation:
44.5x = 0.450 * 9.8
Solving for x, we find that the spring will be stretched by approximately 0.10 m (or 10 cm).