A steel spring is hung vertically. Using a ruler, the lower end of the spring was measured to be 25.0cm from its upper end. However, when an object of mass 0.500kg it attracted to the lower end of the spring the spring stretches to a new length of 45.0cm.
With the aid of a diagram and with a detailed explanation of your reasoning determine.
(a) The stiffness constant k of the spring
(b) The additional weight needed to extend the spring to 50.0cm
(c) The force needed to extend the spring by 50.0mm from its natural length
To find the stiffness constant (k) of the spring, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. Mathematically, it can be expressed as:
F = -kx
Where F is the force applied, k is the stiffness constant, and x is the displacement from the equilibrium position.
(a) To find the stiffness constant (k) of the spring, we need to determine the force (F) applied when the spring is stretched to a new length of 45.0 cm.
First, let's convert the measurements to meters:
Initial length of the spring = 25.0 cm = 0.25 m
New length of the spring = 45.0 cm = 0.45 m
The displacement (x) of the spring is the difference between the new length and the initial length:
x = 0.45 m - 0.25 m = 0.20 m
Now, we can use the weight of the object to determine the force applied to the spring. The weight is given by:
Weight = mass * gravitational acceleration
Given that the object's mass is 0.500 kg and the gravitational acceleration is 9.8 m/s^2, the weight is:
Weight = 0.500 kg * 9.8 m/s^2 = 4.9 N
Since the force exerted by the spring is equal to the weight of the object, we can substitute these values into Hooke's law:
4.9 N = -k * 0.20 m
Now, solve for the stiffness constant (k):
k = -4.9 N / 0.20 m = -24.5 N/m
Therefore, the stiffness constant (k) of the spring is 24.5 N/m.
(b) To find the additional weight needed to extend the spring to 50.0 cm, we need to determine the force required to stretch the spring from 45.0 cm to 50.0 cm.
Again, let's convert the measurements to meters:
New length of the spring = 50.0 cm = 0.50 m
Previous length of the spring = 45.0 cm = 0.45 m
The displacement (x) of the spring is the difference between the new length and the previous length:
x = 0.50 m - 0.45 m = 0.05 m
Using Hooke's law, we can calculate the force required:
F = - k * x
F = -24.5 N/m * 0.05 m = -1.225 N
Since the force exerted by the spring is equal to the weight needed, we need to convert this force to weight. We can use the formula:
Weight = mass * gravitational acceleration
Let's assume the gravitational acceleration is still 9.8 m/s^2. Rearrange the formula:
Weight = F / gravitational acceleration
Weight = -1.225 N / 9.8 m/s^2 = -0.125 kg
To find the additional weight needed, we can take the magnitude of the weight:
Additional weight = |Weight| = |-0.125 kg| = 0.125 kg
Therefore, the additional weight needed to extend the spring to 50.0 cm is 0.125 kg.
(c) To find the force needed to extend the spring by 50.0 mm from its natural length, we need to convert 50.0 mm to meters.
50.0 mm = 50.0/1000 m = 0.050 m
Using Hooke's law, we can calculate the force required:
F = - k * x
F = -24.5 N/m * 0.050 m = -1.225 N
Therefore, the force needed to extend the spring by 50.0 mm from its natural length is 1.225 N.