Question

To understand the use of Hooke's law for a spring.

Hooke's law states that the restoring force F⃗  on a spring when it has been stretched or compressed is proportional to the displacement x⃗  of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed.

Recall that F⃗ ∝x⃗  means that F⃗  is equal to a constant times x⃗ . For a spring, the proportionality constant is called the spring constant and denoted by k. The spring constant is a property of the spring and must be measured experimentally. The larger the value of k, the stiffer the spring.

In equation form, Hooke's law can be written

F⃗ =−kx⃗ .

The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by F=kx, where x is the magnitude of the displacement.

A 60 kg driver gets into an empty taptap to start the day's work.

A 60 kg driver gets into an empty taptap to start the day's work. The springs compress 2.1×10⁻² m . What is the effective spring constant of the spring system in the taptap? The springs compress 2.1×10⁻² m <---- ANSWER

k =

2.8×104

N/m

how do I do part B=

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 60 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

PART -A Because F = k x Because x = compression = 2.1 times 10^- 2 m Weight of the driver = W = mg W = 60 times 9.8 = 588 N Here weight is the force on the spring. W = K x K = W/x = 588/2.1 times 10^- 2 = 2.80 times 10^4 N/m Part - B Force on the spring is equal to sum of all the weight of the individuals. Total mass = M = 27 person + 2 goats + 5 chicken + bananas M = (27 times 60 + 3 times 15 + 5 times 3 + 25) kg M = 1620 + 45 + 15 + 25 M = 17.5 kg So, weight = W_1 = 1705 times 9.8 = 16709 N \r\n and spring constant = k = 2.80 times 10^4 N/m Compression = x = W_1/R x = 16709/2.80