After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average mass of 69kg 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?

We will define the Total mass to calculate the force, so our values are:

Total Mass

The Weight is,

Through the hook's Law we calculate X.

, where x is the lenght of compression and K the Spring constant.

We don't have a K-Spring, but we can assume a random value (or simply let the equation in function of K)

I assume a value of

The compression of the spring at the given weight is 79.5 cm.

"Your question is not complete, it seems to be missing the following information below";

the spring constant of the taptap is k =2.4 × 10⁴ N/m

The given parameters;

mass of each of the 27 persons, = 69 kgmass of each of the 3 goats, = 15 kgmass of each of the 5 chickens, = 3 kgmass of the bananas, = 25 kg

The total mass of all the passengers contained in the taptap is calculated as;

total mass of the persons = 27 x 69 = 1863 kgtotal mass of the goats = 3 x 15 = 45 kgtotal mass of the chickens = 5 x 3 = 15 kgtotal mass of the bananas = 25 kg

The total mass = (1863 + 45 + 15 + 25)kg

The total mass = 1948 kg

The extension of the spring is calculated as follows;

Thus, the compression of the spring at the given weight is 79.5 cm.

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