A spring with k = 53 N/m hangs vertically next to the 15 cm mark on the ruler. If a 2.5 kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks
The added load is F = M g = 24.5 N
The spring will stretch by an amount
delta L = F/k = 0.46 m (46 cm)
That will increase the length to the 46 + 15 = 61 cm mark.
Well, let's see if I can "spring" up with an answer for you! The displacement of the end of the spring can be calculated using Hooke's Law. According to Hooke's Law, the force exerted by the spring is directly proportional to its displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
Given that the spring constant, k, is 53 N/m and the mass, m, is 2.5 kg, we can find the displacement, x, using the formula:
F = kx
Since the mass is hanging vertically, the force exerted by the spring, F, is equal to the weight of the mass, mg:
F = mg
Substituting the values, we have:
mg = kx
2.5 kg * 9.8 m/s^2 = 53 N/m * x
24.5 N = 53 N/m * x
Now, we can solve for x:
x = 24.5 N / (53 N/m)
x ≈ 0.462 m
So, the end of the spring will line up with the ruler marks approximately at the 46.2 cm mark. Hope that "springs" in some laughter while answering your question!
To determine where the end of the spring will line up with the ruler marks, we need to calculate the extension of the spring caused by the 2.5 kg mass attached to it.
The extension of a spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension:
F = -kx
Where:
F is the force applied to the spring (in Newtons),
k is the spring constant (in N/m),
x is the extension of the spring (in meters).
Rearranging the formula, we have:
x = -F / k
Given that the spring constant, k, is 53 N/m, and the force applied to the spring, F, is the weight of the mass, which can be calculated as F = mg (mass times acceleration due to gravity), we can plug in the values:
F = (2.5 kg) * (9.8 m/s^2) = 24.5 N
Substituting the values into the formula, we have:
x = -24.5 N / 53 N/m
x = -0.4623 m
The negative sign indicates that the spring is being extended downward.
To find where the end of the spring will line up with the ruler marks, we need to add the extension (0.4623 m) to the original length of the spring (15 cm or 0.15 m):
Final position = 0.15 m + 0.4623 m
Final position = 0.6123 m
Therefore, the end of the spring will line up with the ruler mark at approximately 0.6123 m (or 61.23 cm).
To find where the end of the spring will line up with the ruler marks, we can use Hooke's Law, which states that the displacement of a spring is directly proportional to the force applied to it.
According to Hooke's Law, the equation to determine the displacement of the spring is:
F = k * x
where F is the force applied to the spring, k is the spring constant, and x is the displacement.
In this case, the force applied to the spring (F) is the weight of the mass attached to it. The weight (W) can be determined using the equation:
W = m * g
where m is the mass (2.5 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the force applied to the spring would be:
F = 2.5 kg * 9.8 m/s^2
Next, we can substitute the value of the force into Hooke's Law equation and solve for the displacement (x):
2.5 kg * 9.8 m/s^2 = 53 N/m * x
Dividing both sides of the equation by 53 N/m:
(2.5 kg * 9.8 m/s^2) / 53 N/m = x
Approximately:
0.464 m = x
Now, we can determine where the end of the spring will line up with the ruler marks. Since the spring hangs vertically next to the 15 cm mark, we need to add the displacement value to the starting position of the spring. Converting the displacement to centimeters (1 m = 100 cm):
0.464 m * 100 cm/m = 46.4 cm
Therefore, the end of the spring will line up with the 46.4 cm mark on the ruler.