An ideal spring with spring constant k is stretched Δx = 0.65 m when a beehive of mass 26 kg is suspended from it as shown in the diagram below. How much will the spring extend if a 62 kg beehive is suspended from the same spring?
To find out how much the spring will extend when a 62 kg beehive is suspended from it, we can use Hooke's Law, which states that the force needed to stretch or compress a spring is directly proportional to the displacement.
Hooke's Law can be expressed as follows: F = -kx
Where:
F is the force applied on the spring,
k is the spring constant,
x is the displacement (change in length) of the spring.
In the first scenario, a beehive of mass 26 kg is suspended from the spring and the spring extends by Δx = 0.65 m.
To find the spring constant, we can rearrange Hooke's Law: k = -F / x
The force applied on the spring can be calculated by multiplying the mass (m) by the acceleration due to gravity (g): F = mg
Substituting the values into the formula, we get:
F = (26 kg) * (9.8 m/s^2) = 254.8 N
Using Hooke's Law, we can find the spring constant:
k = -F / x = -254.8 N / 0.65 m = -392 N/m
Now, we can use the spring constant to find out how much the spring will extend when a 62 kg beehive is suspended from it.
We rearrange Hooke's Law to solve for x:
x = -F / k
Again, we calculate the force applied on the spring:
F = mg = (62 kg) * (9.8 m/s^2) = 607.6 N
Substituting the values into the formula, we get:
x = -F / k = -607.6 N / -392 N/m = 1.55 m
Therefore, when a 62 kg beehive is suspended from the same spring, the spring will extend by approximately 1.55 m.