Question

After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average mass of 66 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? Enter the compression numerically in meters using two significant figures.

Answer 1

Incomplete Question

Explanation:

String constant is not stated

I'll solve this in a general way.

At the end of my explanation, I'll assume any value for string constant.

The total weight on the taptap : Weight of people + Weight of goats + Weight of chickens + weight of banana

= (25 * 66) + (3 * 15) + (5 * 3) + 25 = 1735 kg

The gravitational force exerted on the taptap

Fig = m*g

where m = the total mass 1735kg

and g = 9.8m/s²

Hooke's law of Elasticity states that

Fs = Kx

Where Fs = spring force

k = spring constant

x = spring stretch or compression

Since Fs = Fg

So, Kx = mg

Make x the subject of the formula

x = mg/k --------------- This is the general way of solving for spring compressor in cases like this

I'll replace m and g with their values, respectively.

Because, the spring constant (k) is omitted in the question, I'll assume it to be 30000

x = 1735 * 9.8/30000

x = 17003/30000

x = 0.566767 metres

x = 0.57 metres ------------- Approximated

In conclusion, the spring compression is 0.57 metres if the spring constant is 30000