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In a control system, an accelerometer consists of a 4.63-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 0.832g, the object should be at a location 0.450 cm away from its equilibrium position.

Required:
Find the force constant of the spring required for the calibration to be correct.

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Answer:

8.4 N/m

Explanation:

m = Mass of block = 4.63 gm

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

x = Displacement of spring = 0.45 cm

a = Acceleration of subject = 0.832g

k = Spring constant

Force is given by

[tex]F=ma[/tex]

From Hooke's law

[tex]F=kx[/tex]

So

[tex]ma=kx\\\Rightarrow k=\dfrac{ma}{x}\\\Rightarrow k=\dfrac{4.63\times 10^{-3}\times 0.832\times 9.81}{0.45\times 10^{-2}}\\\Rightarrow k=8.4\ \text{N/m}[/tex]

The force constant of the spring is 8.4 N/m.

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