A69 kg driver gets into an empty taptap to start the day's work. the springs compress 2×10−2 m . what is the effective spring constant of the spring system in the taptap? enter the spring constant numerically in newtons per meter using two significant figures.

Explanation:

The spring system in the taptap obey's Hooke's law, which states that:

where

F is the magnitude of the force applied

k is the spring constant

x is the compression/stretching of the spring

In this problem:

- The force applied is the weight of the driver of mass m = 69 kg, so

- The compression of the spring is

So, the spring constant is

The spring constant of the spring compressed 2x10⁻² m by a 69 kg driver is 3.4x10⁴ N/m.    

We can use the Hooke's law equation to find the spring constant:

  (1)  

Where:

F: is the Hooke's force

k: is the spring constant =?

x: is the distance of compression of the spring = -2x10⁻² m. The negative sign is because it is compressing (negative direction).

The force of equation (1) is equal to the weight force (W) of the driver, so:

  (2)

Where:

m: is the mass of the driver = 69 kg

g: is the acceleration due to gravity = 9.81 m/s²

Solving equation (2) for k, we have:

Therefore, the spring constant is 3.4x10⁴ N/m.

You can learn more about Hooke's law here:

I hope it helps you!                  

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