Respuesta :
Answer:
a) v= 284.44[tex]\frac{m}{s^{2}}[/tex]
b) a=17.78 [tex]\frac{m}{s^{2} }[/tex]
c) x=6.67m
Explanation:
a).
[tex]U(x)=-\alpha *x^{2} +\beta*x^{3}\\\alpha=2\frac{J}{m^{2}}\\\beta=0.3\frac{J}{m^{2}}\\U(0)+V(0)=U(4)+V(4)\\U(0)=-\alpha *0^{2} +\beta*0^{3}=0\\U(0)=0\\V(0)=0\\0=U(4)+V(4)\\U(4)=-\alpha *4^{2} +\beta*4^{3}\\U(4)=-2*4^{2}+0.3*(4^{3})\\U(4)=-12.8 J\\0=-12.8J+V(4)\\12.8=\frac{1}{2}*m*(v_{4})^{2} \\ v_{4}^{2} =\frac{2*12.8J}{0.09kg}\\ v_{4}^{2}=284.44[/tex]
[tex]v_{4}=\sqrt{284.44}\\v_{4}=16.8 \frac{m}{s}[/tex]
b).
[tex]F_{x}=\frac{dU}{dt}\\F_{x}=2*\alpha*x-3*\beta *x^{2} \\F_{4}=2*2*4-3*0.3*(4)^{2}\\F_{4}=1.6N\\F_{4}=m*a\\a=\frac{F_{4}}{m}=\frac{1.6N}{0.09kg}\\a=17.7 \frac{m}{s^{2} }[/tex]
c).
[tex]F_{x}=m*ax\\ax=\frac{F_{x}}{m} \\ax=\frac{4x-0.9x^{2}}{0.09kg}\\\frac{dVx}{dt}= \frac{4x-0.9x^{2}}{0.09kg}\\\int\limits^x_x {\frac{1}{0.09}*(4x-0.9x^{2}) } \, dx\\ \frac{Vx^{2} }{2}=22.2x^{2} -3.3x^{3}[/tex]
Value x,0
[tex]Vx^{2} =44.4x^{2} -6.6x^{3}\\ Vx=\sqrt{44.4x^{2} -6.6x^{3}}[/tex]
Inside the square root is the value of maximum value of x
[tex]44.4x^{2} -6.6x^{3}=0\\x^{2}(44.4-6.6x)=0\\ x=0[/tex] but that value is not real so:
[tex]44.4-6.6x=0\\6.6x=44.4\\x=\frac{44.4}{6.6}\\ x=6.67m[/tex]
The conservation of mechanical energy allows to find the results for the questions of the motion of mass in a conservative force are;
a) the velocity at x = 4 m is: v = 16.9 m / s
b) The acceleration at x = 4m is: a = 17.8 m / s²
c) The maximum elongations is: x = 6.67 m
Given parameters
- The potential energy U = - α x² + β x³ with α= 2 j/m² and β= 0.3 j/m³
- The mass of the body m = 0.0900 kg
- The distance x = 4.00 m
To find
a) The speed
b) The acceleration
c) The maximum value of the distance
The conservation of mechanical energy is one of the most important concepts in physics, stable that if there is not friction the mechanical energy remains constant at all points.
Em = K + U
Where Em mechanical energy, K the kinetical energy ang U the potential energy.
a) Let's find the velocity using the conservation of mechanical energy
Starting point. Where the mass is released.
Em₀ = U (0)
Final point. When for a distance of x = 4 m here we have potential and kinetic energy.
[tex]Em_f[/tex] = K + U (4)
They indicate that the only force is conservative, therefore mechanical energy is conserved
Em₀ = [tex]Em_f[/tex]
0 = ½ m v² + U (4)
½ m v² = -U (4)
v² = 2 / m (αx² - β x³)
Let's calculate
v² = [tex]\frac{2}{0.09}[/tex] (2 4² - 0.3 4³)
v = [tex]\sqrt{284.44}[/tex]
v = 16.9 m / s
b) Acceleration is requested at this point.
We use that potential energy and force are related
F = [tex]- \frac{ dU}{dx}[/tex]
We carry out the derivatives
F = 2αx - 3βx²
Let's calculate
F = 2 2 4 - 3 0.3 4²
F = 1.6 N
Now we use Newton's second law that relates the net force with the product of the mass and the acceleration of the body.
F = ma
a =[tex]\frac{F}{m}[/tex]
a = [tex]\frac{1.6}{0.09}[/tex]
a = 17.8 m / s²
c) At maximum displacement.
Let's use conservation of mechanical energy
Starting point. Where x = 0 is released
Emo = U (0) = 0
Final point. Point of maximum elongation, kinetic energy is zero
[tex]Em_f[/tex] = U (xmax)
Energy is conserved
Em₀ = [tex]Em_f[/tex]
[tex]0 = U(x_{max})[/tex]
-αx² + βx³ = 0
x² (-α + βx) = 0
the solutions of this equation is:
x = 0
-α + βx = 0
x = [tex]\frac{\alpha}{\beta }[/tex]
Let's calculate
x = [tex]\frac{2}{0.3}[/tex]
x = 6.67 m
In conclusion using the conservation of mechanical energy we can find the results for the questions of the motion of mass in a conservative force are;
a) the velocity at x = 4 m is: v = 16.9 m / s
b) The acceleration at x = 4m is: a = 17.8 m / s²
c) The maximum elongations is: x = 6.67 m
Learn more here: brainly.com/question/2615468
