Question

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. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 0.800g, the object should be at a location 0.500 cm away from its equilibrium position. Find the force constant of the spring required for the calibration to be correct.

Answer 1

The force constant of the spring required for the calibration to be correct is approximately 0.0362 N/m.

Step-by-step explanation:

To find the force constant of the spring required for the calibration to be correct, we can use the equation for Hooke's Law: F = -kx, where F is the force, k is the force constant of the spring, and x is the displacement from the equilibrium position. In this case, the force is given by the product of the mass and acceleration: F = ma. We can substitute this into the equation for Hooke's Law and solve for k:

k = -ma / x

Given that the mass is 4.70 g (or 0.00470 kg), the acceleration is 0.800g (or 0.800 * 9.8 m/s^2), and the displacement is 0.500 cm (or 0.00500 m), we can plug these values into the equation to find the force constant of the spring.

Using the equation: k = -(0.00470 kg * 0.800 * 9.8 m/s^2) / 0.00500 m

k = -0.0362 N/m

Therefore, the force constant of the spring required for the calibration to be correct is approximately 0.0362 N/m.

Answer 2

7.37 N/m

Step-by-step explanation:

Given:

Mass of the object = 4.70 g = 0.0047 kg

acceleration = 0.800 g = 0.800 × 9.81 = 7.848 m/s²

Displacement = 0.500 cm = 0.005 m

Now,

Force by the object = Mass × Acceleration

= 0.0047 × 7.848

= 0.0368856 N

and,

Force by the spring = kx

where, x is the displacement

k is the spring constant

thus,

0.0368856 = k × 0.005

or

k = 7.37 N/m