In a control system, an accelerometer consists of a 5.15-g object sliding on a calibrated horizontal rail. A low-mass spring attaches the object to 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 0.864g, the object should be at a location 0.465 cm away from its equilibrium position. Find the force constant of the spring required for the calibration to be correct
In a control system, an accelerometer consists of a
4.70-g object sliding on a calibrated horizontal rail. A
low-mass spring attaches the object to a flange at one
end of the rail.
the force on the object is
F = ma = 0.00511 kg*0.824 m/s^2 = 0.00421 N
and k = N/s = 0.00421 N/0.0455 m = 0.09254 N/m
To find the force constant of the spring required for the calibration to be correct, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Hooke's Law can be written as:
F = -kx
where:
F is the force exerted by the spring,
k is the force constant (in N/m),
x is the displacement from the equilibrium position.
In this case, the steady acceleration acting on the object is given as 0.864g, where g represents the acceleration due to gravity.
We know that the object should be at a location 0.465 cm away from its equilibrium position when subjected to this steady acceleration.
Let's go step-by-step to find the force constant (k):
Step 1: Convert the acceleration to m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
Steady acceleration = 0.864g = 0.864 * 9.8 m/s^2
Step 2: Convert the displacement to meters
Displacement = 0.465 cm = 0.465/100 m
Step 3: Substitute the values into Hooke's Law equation
F = -kx
0.864 * 9.8 = -k * (0.465/100)
Step 4: Solve for k
k = -(0.864 * 9.8) / (0.465/100)
Simplifying the expression:
k ≈ -17.86 N/m
Therefore, the force constant of the spring required for the calibration to be correct is approximately 17.86 N/m.
To find the force constant of the spring required for the calibration to be correct, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is given as:
F = -k * x
Where:
F is the force exerted by the spring,
k is the force constant of the spring,
x is the displacement from the equilibrium position.
In this case, we are given:
Acceleration, a = 0.864g
Displacement, x = 0.465 cm
Mass, m = 5.15 g
First, convert the mass from grams to kilograms:
m = 5.15 g / 1000 = 0.00515 kg
Next, calculate the force exerted on the object due to acceleration:
F = m * a
F = 0.00515 kg * 0.864g
F = 0.004454 kg·m/s²
Now, substitute the known values into Hooke's Law and solve for k:
0.004454 kg·m/s² = -k * 0.00465 m
To isolate k, divide both sides of the equation by 0.00465:
k = 0.004454 kg·m/s² / 0.00465 m
Simplifying the expression:
k ≈ 0.957 kg/s²
Therefore, the force constant of the spring required for the calibration to be correct is approximately 0.957 kg/s².