When a 2.50-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.13 cm.
a)What is the force constant of the spring?
b) If the 2.50-kg object is removed, how far will the spring stretch if a 1.25-kg block is hung on it?
k= force/x= mg/.0313
b) x= force/k=mg/k it better be half as far.
To find the force constant of the spring, we can use Hooke's law formula. Hooke's law states that the force exerted by a spring is directly proportional to the displacement or stretching of the spring.
a) Force constant (k) can be calculated using the formula:
k = F / x
Where:
- k is the force constant of the spring
- F is the force applied to the spring
- x is the displacement or stretching of the spring
In this case, we know that the mass of the object hanging on the spring is 2.50 kg, and the spring stretches 3.13 cm (which we'll convert to meters). We need to find the force applied to the spring.
First, let's find the force applied to the spring:
F = m * g
Where:
- m is the mass of the object (2.50 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
F = 2.50 kg * 9.8 m/s^2
F ≈ 24.5 N (rounded to two decimal places)
Now let's convert the displacement to meters:
3.13 cm = 0.0313 m
Now we can calculate the force constant:
k = 24.5 N / 0.0313 m
k ≈ 782.34 N/m (rounded to two decimal places)
Therefore, the force constant of the spring is approximately 782.34 N/m.
b) To determine how far the spring will stretch when a 1.25-kg block is hung on it, we can use the same formula as before:
F = k * x
Where:
- F is the force applied to the spring (which we'll calculate)
- k is the force constant of the spring (782.34 N/m as calculated previously)
- x is the displacement or stretching of the spring (which we need to find)
First, let's calculate the force applied to the spring:
F = m * g
Where:
- m is the mass of the block (1.25 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
F = 1.25 kg * 9.8 m/s^2
F ≈ 12.25 N (rounded to two decimal places)
Now we can rearrange the formula to find x:
x = F / k
x = 12.25 N / 782.34 N/m
x ≈ 0.0156 m (rounded to four decimal places)
Therefore, when the 1.25-kg block is hung on the spring, the spring will stretch approximately 0.0156 meters.
a) To find the force constant of the spring, we can use Hooke's law, which states that the force exerted by a spring is proportional to its displacement from its equilibrium position.
Hooke's law can be written as F = kx, where F is the force applied, k is the force constant (also known as spring constant), and x is the displacement.
We are given that the object weighs 2.50 kg and the spring stretches 3.13 cm (or 0.0313 m).
Therefore, the force applied (F) can be calculated using the formula F = mg, where m is the mass of the object (2.50 kg) and g is the acceleration due to gravity (9.8 m/s^2).
F = (2.50 kg)(9.8 m/s^2) = 24.5 N
We can substitute the values for F and x into Hooke's law to find the force constant (k):
24.5 N = k(0.0313 m)
Divide both sides by 0.0313 m:
k = 24.5 N / 0.0313 m ≈ 781.25 N/m
So, the force constant of the spring is approximately 781.25 N/m.
b) To find how far the spring will stretch when a 1.25-kg block is hung on it, we can once again use Hooke's law.
We know the mass of the object (1.25 kg) and the force constant of the spring (781.25 N/m). Now, we need to find the force applied first.
The force applied (F) can be calculated using the formula F = mg:
F = (1.25 kg)(9.8 m/s^2) = 12.25 N
Now, we can substitute the values for F and k into Hooke's law to find the displacement (x):
12.25 N = (781.25 N/m)(x)
Divide both sides by 781.25 N/m:
x = 12.25 N / 781.25 N/m ≈ 0.0157 m
So, the spring will stretch approximately 0.0157 m when a 1.25-kg block is hung on it.