Physics, 30.06.2019 05:20 Dweath50

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.

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Answer from: ytcampos6

$3.4\cdot 10^4 N/m$

Explanation:

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

F=kx

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

F=mg=(69 kg)(9.8 m/s^2)=676.2 N

- The compression of the spring is

$x=2\cdot 10^{-2} m=0.02 m$

So, the spring constant is

$k=\frac{F}{x}=\frac{676.2 N}{0.02 m}=3.4\cdot 10^4 N/m$

Answer from: brutusjohnson1

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:

F = -kx (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:

W = F

mg = -kx (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:

$k = -\frac{mg}{x} = -\frac{69 kg*9.81 m/s^{2}}{-2\cdot 10^{-2} m} = 3.4 \cdot 10^{4} N/m$

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

You can learn more about Hooke's law here:

linklink

I hope it helps you!

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