A spring is suspended from an elevator, and a 13.0kg mass is then attached to the spring. When the elevator moves down at an acceleration of 2.00m/s2 , the net elongation of the spring is 25.0cm . Find the spring constant.
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount of elongation or compression of the spring.
Hooke's Law can be written as F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the elongation or compression of the spring.
In this case, the force exerted by the spring is equal to the weight of the mass attached to it. The weight can be calculated using the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. However, since the elevator is moving downwards, we need to take into account the apparent weight, which is reduced by the acceleration of the elevator.
The apparent weight can be calculated using the formula W' = m(g - a), where W' is the apparent weight, m is the mass, g is the acceleration due to gravity, and a is the acceleration of the elevator.
Now, let's calculate the force exerted by the spring:
F = W' = m(g - a)
Next, we need to convert the elongation of the spring from centimeters to meters:
x = 25.0 cm = 0.25 m
Finally, we can substitute the values into Hooke's Law and solve for the spring constant:
k = F/x = (m(g - a))/x = (13.0 kg * (9.8 m/s^2 - 2.00 m/s^2))/(0.25 m)
k ≈ (13.0 kg * 7.8 m/s^2)/(0.25 m)
k ≈ 405.6 N/m
Therefore, the spring constant is approximately 405.6 N/m.