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certain coiled wire with uneven windings has the property that to stretch it an amount s from its relaxed length requires a force that is given by, $F=b{s}^3$so its behaviour is different from a normal spring. You suspend this device vertically, and its unstretched length is 25 cm. (a) You hang a mass of 18 g from the device, and you observe that the length is now 29 cm. What is b, including units? (b) Which of the following were needed in your analysis in part (a)? (1) The Momentum Principle, (2) The fact that the gravitational force acting on an object near the Earth’s surface is approximately mg, (3) The force law for an ordinary spring ($F=k_{s}s$), (4) The rate of change of momentum being zero (c) Next you take hold of the hanging 18 g mass and throw it straight downward, releasing it when the length of the device is 33 cm and the speed of the mass is 5 m/s. After a very short time, 0.0001 s later, what is the stretch of the device, and what was the change in the speed of the mass (including the correct sign of the change) during this short time interval? It helps enormously to draw a diagram showing the forces that act on the mass after it leaves your hand.

Step by step solution

01

Identification of the given data

The given data can be listed below as-

• The upstretched length of the coil is 25 cm.
• The mass of the device is18 g.
• The stretched length of the coil after 18g load applied is 29cm

02

Significance of the Newton’s first law

The Newton’s first law states that a body will continue to move in uniform motion unless it is acted by an external force.

The equation of motion gives the force and the diagram.

03

Determine the balance force

a)

the equation of the force given in the question is,

$$\begin{gathered} F=b{s}^3 \\ =b{\left(s_f-s_i\right)}^3 \end{gathered}$$

Here, F is the acting force and s is the difference between the final and the initial length.

Substituting the values in the above equation,

$$\begin{gathered} b{s}^3=mg \\ b{\left(0.29-25m\right)}^3=18×{10}^{-3}kg×9.8m/s^2 \\ b=\frac{0.1764kgm/s^2}{{\left(0.04m\right)}^3} \\ =2.75×{10}^{3}kg/m^2{s}^2×\frac{1N/m^3}{1g/m^2{s}^2} \end{gathered}$$

Hence, further as,

$b=2.75×{10}^{3}N/m^3$

Thus, the value of b is$2.75×{10}^{3}N/m^3$.

04

What is needed in analysis in part (a)

b)

As the above solution was based on the concept of force, then the concept is needed is F = mg

Thus, 2) was needed in the analysis.

05

Step 5:Determination of the stretch of the device and the change in the speed of the mass along with the diagram

c)

The equation of the acceleration is expressed as,

$a=\frac{b{s}^3-mg}{m}$

Here, a is the acceleration, m is the mass, b is the constant and s is the stretch.

Substituting the values in the above equation,

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