Final answer:
To calculate the effective spring constant of the spring system in a taptap, use Hooke's Law with the driver's weight as the force causing spring compression. The calculation results in an effective spring constant of 25480 N/m.
Step-by-step explanation:
The problem given involves a 65 kg driver getting into an empty taptap and the springs compress by 2.5×10⁻² m. To determine the effective spring constant (k) of the spring system, we can use Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, expressed as F = kx. In this scenario, the force exerted on the spring is due to the weight of the driver, which can be calculated as F = mg, where m is the mass of the driver and g is the acceleration due to gravity (approximately 9.8 m/s²).
We can set up the equation mg = kx and solve for k:
- F = mg = 65 kg × 9.8 m/s² = 637 N (Newtons)
- x = 2.5×10⁻² m
- k = F/x = 637 N / 2.5×10⁻² m = 25480 N/m
Therefore, the effective spring constant of the spring system in the taptap is 25480 N/m.