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4 years ago

A non-relativistic particle of mass m moves in one dimension x under the force

Physics

1 answer:

marin [14]4 years ago

6 0

Answer:

U = - (x⁴ / 4 - b x² / 2) , c) The function is zero for x = 0 and √2

Explanation:

a and b) Strength and potential energy are related

F = - dU / dx

Therefore to find the energy we must integrate

∫ dU = -∫ F dx

∫ dU = - ∫ (a x³ –b x) dx

Let's make the integration

U = - (x⁴ / 4 - b x² / 2)

We evaluate the integral between the value

U - U₀ = -x⁴ / 4 + x² / 2 - (-x₀⁴ / 4 + x₀² / 2)

The arbitrary constant is zero, so that U is zero in the zero position

U₀ = 0 for x₀ = 0

c) Mechanical energy is the sum of kinetic energy plus potential energy

Em = K + U

Em = ½ m v² + ½ (x² -x⁴ / 2)

E = ½ m v² + ½ x² (1 - x² / 2)

Energy is positive

2 (E –K) = x² (1-x² / 2)

At the return points K = 0

The zero points of this function are

x = 0

(1- x² / 2) = 0

x² = 2

x = √ 2

The function is zero

x = 0 and √2

d) the movement is bounded for energy values less than or equal to

E <= ½ x² (1-x² / 2)

e) for this part we resolved Newton's second law

F = m a

ax³ - b x = m d²x / dt²

d²x / dt² = -b / m x + a / m x³3

The linear term gives a simple harmonic movement

w₀² = b / m

d²x / dt² = - w₀² x + a / m x³

The frequencies are the frequencies of the harmonic movement plus a small change due to the non-harmonic part of the movement

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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$