Question

Learning Goal:

To understand the use of Hooke's law for a spring.
Hooke's law states that the restoring force F on a spring when it has been stretched or compressed is proportional to the displacement x of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed.
Recall that F∝x means that F is equal to a constant times x . 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
F =−kx .
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.
In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs.
Part A
A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck.
A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.4×10−2 m . What is the effective spring constant of the spring system in the taptap?
2.9x10^4
Part B
After driving a portion of the route, the taptap is fully loaded with a total of 23 people including the driver, with an average mass of 70 kg per person. In addition, there are three 15-kg goats, five 3-kgchickens, 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?
Part C
Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic?

Answer 1

A) k = 2,858 10⁴ N / m , B) x_total = - 5,812 10⁻¹ m

C) it is very possible that the obtained value is not realistic for small vehicles

Step-by-step explanation:

Part A

In this case we can use Hooke's law

F = - k x

force is the weight of the driver

F = W

mg = - k x

k = - mg / x

the springs are compressed x = - 2,4 10⁻² m

k = - 70 9.8 / (-2.4 10⁻²)

k = 2,858 10⁴ N / m

Part B

Since we have the spring constant we must find the complete weight, for this we look for the total masses

each mass is the number of element by the mass of an element

M = 23 70 + 3 15 + 5 3 + 1 25

M = 1695 kg

F = -k x

F = W = M g

Mg = - k x_total

x_total = -M g / k

x_total = -1695 9.8 / 2,858 10⁴

x_total = - 5,812 10⁻¹ m

The negative sign indicates that the springs are compressing

Part C

The truck has lowered 0.58 m = 58 cm

This drop is very large probably in a real vehicle with this drop it would be touching the ground or very close, therefore it is very possible that the obtained value is not realistic for small vehicles