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 attached 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.
M*g = 0.5 * 9.8 = 4.9 N. = Wt. of object.
a. k = F/d = 4.9N./(0.45-0.25)m = 24.5 N/m.
b. 24.5N./m * (0.50-0.25)m = 6.125 N.
6.125 - 4.9 = 1.23 N. addition wt.
c. F = 24.5N/m * 0.05m = 1.23 N.
Henry, the problem I am trying to solve here is the explanation part. How am I meant to provide such a detailed explanation when the majority of the methods make up the decription of the answer?
To answer these questions, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. Mathematically, Hooke's Law is expressed as F = -kx, where F is the force exerted by the spring, k is the stiffness constant (also known as the spring constant), and x is the displacement from the equilibrium position.
(a) To find the stiffness constant (k) of the spring, we can use the information given. We know that the spring stretches from 25.0cm to 45.0cm when an object with a mass of 0.500kg is attached to it.
First, let's convert the measurements to meters:
Initial length (L1) = 25.0cm = 0.25m
Final length (L2) = 45.0cm = 0.45m
Next, we need to find the change in length or displacement (x):
Displacement (Δx) = L2 - L1 = 0.45m - 0.25m = 0.20m
The force exerted by the spring can be calculated using Hooke's Law:
F = -kx
We need to find F, so we'll rearrange the equation:
k = -F / x
The force (F) can be determined using the equation F = mg, where m is the mass of the object and g is the acceleration due to gravity. Plugging in the values:
F = (0.500kg)(9.8m/s^2) = 4.90N
Now, we can find the stiffness constant (k):
k = -F / x = -4.90N / 0.20m = -24.5 N/m
Therefore, the stiffness constant (k) of the spring is 24.5 N/m.
(b) To determine the additional weight needed to extend the spring to 50.0cm, we can use Hooke's Law again.
The change in length or displacement from the initial position is:
Δx = L2 - L1 = 0.50m - 0.25m = 0.25m
Using Hooke's Law (F = -kx), we need to find the force (F) required to extend the spring by this displacement.
F = -kx = -(24.5 N/m)(0.25m) = -6.125N
To determine the additional weight needed, let's assume it has the same stiffness constant as the original spring.
Using the equation F = mg and rearranging it to solve for mass (m), we have:
m = F / g = (-6.125N) / (9.8m/s^2) ≈ -0.626kg
We have a negative value for mass because the additional weight needs to be placed in the opposite direction of the force exerted by the spring.
Therefore, an additional weight of approximately 0.626kg (or 626g) is needed to extend the spring to 50.0cm.
(c) To find the force needed to extend the spring by 50.0mm (or 0.050m) from its natural length, we can again use Hooke's Law.
The displacement (x) is given as 0.050m.
Using Hooke's Law (F = -kx):
F = -kx = -(24.5 N/m)(0.050m) = -1.225N
Therefore, the force needed to extend the spring by 50.0mm from its natural length is approximately 1.225N.