A certain spring stretches 8.9 cm when it supports a mass of 0.89 kg .
If the elastic limit is not reached, how far will it stretch when it supports a mass of 18 kg ?
To solve this problem, we need to use Hooke's Law, which states that the force exerted by a spring is 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 exerted by the spring,
k is the spring constant,
x is the displacement of the spring.
To find the spring constant (k), we can use the given information. We know that the spring stretches 8.9 cm (0.089 m) when it supports a mass of 0.89 kg. The force exerted by the spring can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2):
F = (0.89 kg) * (9.8 m/s^2) = 8.722 N
Now, we can substitute the known values into Hooke's Law to find the spring constant:
8.722 N = k * 0.089 m
Solving for k, we get:
k ≈ 97.91 N/m
Now that we know the spring constant, we can use Hooke's Law to find how far the spring will stretch when it supports a mass of 18 kg. Again, we can calculate the force exerted by the spring:
F = (18 kg) * (9.8 m/s^2) = 176.4 N
Now, we can rearrange Hooke's Law to solve for x (the displacement of the spring):
x = F / k
Substituting the known values, we have:
x = 176.4 N / 97.91 N/m
Simplifying, we get:
x ≈ 1.8 m
Therefore, the spring will stretch approximately 1.8 meters when it supports a mass of 18 kg.