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A 60 kg driver gets into an empty taptap to start the day's work. The springs compress 2.4×10−2 m . What is the effective spring constant of the spring system in the taptap?

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We can use Hooke's Law to calculate the effective spring constant of the spring system.

Hooke's Law states that the force exerted by a spring is proportional to its extension or compression from its equilibrium position. The formula for Hooke's Law is:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In this problem, the driver has a weight of 60 kg. The force exerted by the spring is equal to the weight of the driver, so we can write:

F = mg = (60 kg)(9.8 m/s^2) = 588 N

The displacement of the spring is given as 2.4×10−2 m.

We can now use Hooke's Law to solve for the spring constant:

k = -F/x = -(588 N)/(2.4×10−2 m) = -24500 N/m

The effective spring constant of the spring system in the taptap is 24500 N/m.
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User Egor Litvinchuk
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