An accelerometer in a control system consists of a 2.42 g object sliding on a horizontal rail. A low-mass spring is connected between the object and a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 3.50 m/s2, the object must be located 0.370 cm from its equilibrium position. Find the force constant required for the spring.
ma=kx
k=ma/x =2.42•10⁻³•3.5/0.37•10⁻²=2.3 N/m
To find the force constant required for the spring in the control system, we can use Hooke's Law.
Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.
The equation for Hooke's Law is given by:
F = k * x
Where:
F is the force applied to the spring,
k is the force constant (also known as the spring constant), and
x is the displacement from the equilibrium position.
In this scenario, the object is subject to a steady acceleration of 3.50 m/s^2 and is located 0.370 cm from its equilibrium position. Since acceleration is related to the force, we can equate the two using Newton's second law:
F = m * a
Where:
F is the force,
m is the mass of the object, and
a is the acceleration.
We are given that the object weighs 2.42 g. To convert it to kilograms, we divide by 1000:
m = 2.42 g / 1000 = 0.00242 kg
Substituting the values into the equation, we have:
F = 0.00242 kg * 3.50 m/s^2 = 0.00847 N
Now we can find the force constant (k) using Hooke's Law. Rearranging the equation, we have:
k = F / x
Substituting the values, we get:
k = 0.00847 N / 0.00370 m = 2.29 N/m
Therefore, the force constant required for the spring in the control system is 2.29 N/m.