Question

1. A 63 kg driver gets into an empty taptap to start the day's work. The springs compress 1.5x10-2 m. What is the effective spring constant of the spring system in the taptap?

2. After driving a portion of the route, the taptap is fully loaded with a total of 24 people including the driver, with an average mass of 68 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?

Answer 1

(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that

∑ F = s - mg = 0 ==> s = mg = 617.4 N

If the spring is compressed 0.015 m, then the spring constant k is such that

617.4 N = k (0.015 m) ==> k = 41,160 N/m ≈ 41 kN/m

(2) The total mass of the passengers is

24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg

so that if everyone is at rest, the spring is compressed a distance x such that

kx = (1717 kg) g ==> x ≈ 0.41 m