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
Forcefromspring=mass*a=kx
k= massobject*a/x=.0242kg*3.5m/s^2 *1/.00370
figure k.
I am not getting the right answer!
To find the force constant required for the spring, we can start by using Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
In this case, the object has a mass of 2.42 g. However, it's better to convert the mass to kilograms for consistency in the SI system. Since 1 g = 0.001 kg, the mass of the object is:
m = 2.42 g * 0.001 kg/g = 0.00242 kg
The acceleration is given as 3.50 m/s². Now we can calculate the force acting on the object:
F = 0.00242 kg * 3.50 m/s² = 0.00847 N
Next, we need to consider the displacement from the equilibrium position. The object must be located 0.370 cm away. Again, it's better to convert this distance to meters:
d = 0.370 cm * 0.01 m/cm = 0.0037 m
The force acting on the object is related to the displacement by Hooke's law:
F = k * d
where k is the force constant of the spring. By rearranging the equation, we can solve for the force constant:
k = F / d = 0.00847 N / 0.0037 m ≈ 2.29 N/m
Therefore, the force constant required for the spring is approximately 2.29 N/m.