A spring scale hung from the ceiling stretches by 6.1 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and replaced with a 2.5 kg mass.
Part A
What is the stretch of the spring?
Ah, spring physics, always bouncing into our lives. So, if a 2.0 kg mass made the spring stretch by 6.1 cm, let's see what happens when we introduce a 2.5 kg mass.
Now, if the 2.0 kg mass made the spring stretch like it's doing a yoga class, we can assume that the stretch is directly proportional to the mass. So, let's do some springy math.
The stretch caused by the 2.0 kg mass is 6.1 cm. If we want to find the stretch caused by the 2.5 kg mass, we can use some good ol' proportions.
Using the magic of proportions, we can set up a ratio:
(2.0 kg / 2.5 kg) = (6.1 cm / x)
Now, all we have to do is solve for x. Get ready for some number crunching!
Cross multiplying gives us:
2.0 kg * x = 2.5 kg * 6.1 cm
After we unlock our inner math wizard and divide both sides by 2.0 kg, we find that:
x = (2.5 kg * 6.1 cm) / 2.0 kg
So, after a bit of arithmetic, the stretch caused by the 2.5 kg mass is approximately the result of that calculation. Give it a whirl and see what you get! And remember, the journey is just as important as the answer. Enjoy!
To find the stretch of the spring, we can use Hooke's law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's law is:
F = k * x
Where:
F is the force applied to the spring.
k is the spring constant.
x is the displacement of the spring.
In this case, we are given the mass of the object attached to the spring and the displacement of the spring.
We can start by finding the force applied to the spring when the 2.0 kg mass is attached.
Step 1: Convert the mass from kilograms to Newtons.
1 kg = 9.8 N (gravitational acceleration)
So, the force applied when the 2.0 kg mass is attached is:
F = 2.0 kg * 9.8 N/kg
F = 19.6 N
Step 2: Use Hooke's law to find the spring constant.
We know that when the 2.0 kg mass is attached, the spring stretches by 6.1 cm (or 0.061 m).
19.6 N = k * 0.061 m
Now we can solve for k:
k = 19.6 N / 0.061 m
k ≈ 320.65 N/m
Step 3: Find the stretch of the spring when the 2.5 kg mass is attached.
We can use Hooke's law again to find the stretch of the spring when the 2.5 kg mass is attached:
F = k * x
We know that the force applied when the 2.5 kg mass is attached is:
F = 2.5 kg * 9.8 N/kg
F = 24.5 N
Now we can solve for x:
24.5 N = 320.65 N/m * x
x = 24.5 N / 320.65 N/m
x ≈ 0.0762 m
Therefore, the stretch of the spring when the 2.5 kg mass is attached is approximately 0.0762 m (or 7.62 cm).
To find the stretch of the spring, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be written as F = k * x, where F is the force applied, k is the spring constant, and x is the displacement.
In this case, we are given the mass hung from the spring (2.0 kg), and the stretch of the spring (6.1 cm or 0.061 m). We know that the force acting on the spring is equal to the weight of the mass, which can be calculated using the formula F = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
1. Calculate the force acting on the spring:
F = m * g
F = 2.0 kg * 9.8 m/s^2
F = 19.6 N
2. Rearrange Hooke's Law equation to solve for the spring constant (k):
F = k * x
k = F / x
k = 19.6 N / 0.061 m
k ≈ 320.79 N/m
Now that we know the spring constant, we can use it to find the stretch of the spring when the 2.5 kg mass is hung.
3. Calculate the stretch of the spring:
F = k * x
x = F / k
x = 19.6 N / 320.79 N/m
x ≈ 0.061 m
Therefore, the stretch of the spring when the 2.5 kg mass is hung is approximately 0.061 m or 6.1 cm.
F = k x
k = 2 g/6.1 cm
2.5g = (2g/6.1cm) x
x = 6.1 (2.5/2) cm