Questions: A 65 kg driver gets into an empty taptap to start the day's work. The springs compress 2.5 × 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.

A 65 kg driver gets into an empty taptap to start the day's work. The springs compress 2.5 × 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.
Transcript text: A 65 kg driver gets into an empty taptap to start the day's work. The springs compress $2.5 \times 10^{-2} \mathrm{~m}$. 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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Solution

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Solution Steps

Step 1: Calculate the force exerted by the driver

The force exerted by the driver due to their weight is equal to their mass multiplied by the acceleration due to gravity (approximately 9.8 m/s²). F = mg = (65 kg)(9.8 m/s²) = 637 N

Step 2: Calculate the spring constant

Hooke's Law states that the force exerted by a spring is proportional to its displacement from equilibrium: F = kx, where k is the spring constant and x is the displacement. We are given that the springs compress by 2.5×10⁻² m, and we just calculated the force. We can solve for k: k = F/x = (637 N) / (2.5×10⁻² m) = 25480 N/m

Step 3: Round to two significant figures

The problem asks for the answer to two significant figures, so we round 25480 to 25000, or 2.5×10⁴ N/m.

Final Answer: The effective spring constant is 2.5×10⁴ N/m

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