Respuesta :
Answer:
e) 11 m/s
Explanation:
For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:
[tex]K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4][/tex]
In [tex]x_1=1.0m[/tex] the speed is given, so [tex]v_1=5.0\frac{m}{s}[/tex] and [tex]x_0=0[/tex]. Replacing:
[tex]\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}[/tex]
The speed of the particle at the origin is 11.18 m/s. Option E shows the speed of the particle at the origin.
What are potential and kinetic energy?
The energy stored within an object due to its stable position is called the potential energy of the object.
The energy stored within an object due to the motion is called the kinetic energy of the object.
Given that the mass m of the particle is 0.20 kg. The particle has a speed of 5.0 m/s when it is at x = 1.0 m.
The potential energy of the particle is given by the function.
[tex]U(x) = 8.0 x^2 + 2.0 x^4[/tex]
Let's consider that there are two points A and B at the coordinate plane. These are at x=0 and x=1 respectively.
The particle moves under a conservative force, its mechanical energy at point A is will be equal to mechanical energy at point B.
[tex]U(x_a) + K(x_a) = U(x_b) + K(x_b)[/tex]
Where K is the kinetic energy of the particle at points A and B.
[tex](8.0x_a^2 + 2.0x_b^4) + \dfrac {1}{2}mv_a^2 = (8.0x_b^2 + 2.0x_b^4) + \dfrac {1}{2}mv_b^2[/tex]
Put the value of x_a = 0, x_b = 1, v_b = 5.0 and m = 0.20,
[tex](8.0 \times 0 + 2.0 \times 0) + \dfrac {1}{2} \times 0.20 \times v_a^2 = (8.0 \times 1 + 2.0 \times 1) + \dfrac {1}{2} \times 0.20 \times 5^2[/tex]
[tex]\dfrac{1}{2}\times0.20 \times v_a^2 = 10 + 2.5[/tex]
[tex]v_a^2 = \dfrac {12.5 \times 2}{0.20}[/tex]
[tex]v_a^2 = 125[/tex]
[tex]v_a = 11.18 \;\rm m/s[/tex]
Hence we can conclude that the speed of the particle at the origin is 11.18 m/s. Option E shows the speed of the particle at the origin.
To know more about the potential and kinetic energy, follow the link given below.
https://brainly.com/question/21288807.
