PLEASE RESPOND TO physics questions

Question

To answer the questions below, it may be useful to think of your friend's car driving on a level road on the surface of the Earth, or maybe in space accelerating upwards at 9.8 m/s² (or some other rate of acceleration, depending on the question).

1) Your friend starts out by hanging is fuzzy dice from a spring. On the surface of the Earth, he finds the length of the spring to be 8.4 cm. With his car drifting in space he finds the length of the spring to be 3.5 cm. What would be the length of the spring in a situation, if the car were accelerating upward at a rate of 9.8 m/s²?

2) What would be the length of the spring in a situation if the car were accelerating upward at a rate of 12.4 m/s²?

3) What would be the length of the spring in a situation if the car were accelerating upward at a rate of 6.6 m/s²?

NOTE: ANSWER ALL 3 questions since they correspond with each other please I need help and be able to respond if wrong

Answer 1

Here, we have conjoint application of Hook's law of elasticity and Newton's law of motion.

Hook's law is based on the stress-strain properties of materials caused by electrostatic binding forces of atoms and molecules of materials. The internal elasticity force is assumed proportional to the stretching length x of the spring, F_s = k.x.

k is the constant of elastic stretching of the spring and must be determined for numerical results, or left variable for general results.

Newton's law is based on the inertial effects of acceleration on matter. A loaded spring resisting external force F_e = m.a.

The logical approach is to equate elastic force with external accelerating effect. i.e.,

F_e = m.a.

F_s = k.x.

k.x. = m.a. -------------------------(1)

Equation (1) is the governing physical law, which of course is hypothetical, based on the linear stress-strain relationship

x ∝ d²x/dt²

That is Newton's second law of motion that assumes that the speed of rate of change of length x represents force. (Remember, we said speed of rate of change, which is another way of saying "acceleration").

Given:

(a) Length of the spring, L = 8.4 cm when the car (object that stretches the spring, or hanging from string) is pulling by its weight of gravity, g = 9.8 m/s².

(b) The stretching length = Length of spring on ground - Length of spring in space (where gravity is assumed null.)

Length of spring is space, L_min = 3.5 cm

Therefore, x = L - L_min = 8.4 - 3.5 = 4.9 cm --------------------(2)

This given fact suffices to determine the constant k of elastic stretching in equation (1).

Determining he constant k of elastic stretching:

k.x. = m.a. -------------------(Elastic Stretch law)

x= 4.9 cm --------------------(Given stretch in centimeters)

g = 9.8 m/s².-----------------(gravitation acceleration g)

On substitution by x and g in equation (1), we get

k . (4.9 cm / 100) = m . 9.8 cm/s². ------------(metric conversion to m, kg, s, N)

Hence,

k / m = 9.8 (100)/ 4.9 = 200 N / (kg.m) -----------------(3)

Recall that both k and m are constants and need not be known, since we are dealing with ratios and proportions.

Vector Analysis:

An car accelerating at rate a m/s² acquires extra force, in addition the gravity, given by

k.x = m. ( g + a ) -------------------------------(4)

Hence, as a = 12.4 m/s² and 6.6 m/s², x becomes:

x = (m/ / k) ( g + a)

--= ( 1/ 200 ) ( 9.8 + 12.4 ) = 0.111 meter ( when a = 12.4 m/s²)

--= ( 1/ 200 ) ( 9.8 + 6.6 ) = 0.082 meter ( when a = 6.6 m/s²)

Since, the unstrained spring has a length L_min = 3.5 cm - 0.035 meter, then the spring would measure:

L _stretched = 0.111 + 0.035 = 0.146 m = 14.6 cm ( when a = 12.4 m/s²)

------------- = 0.082 + 0.035 = 0.117 m = 11.7 cm ( when a = 6.6 m/s²)