Respuesta :
Answer: a) -13.25 N, b) 1.78 m/sec
Step-by-step explanation:
Since we have given that
[tex]U(x)=ax^4+bx^2[/tex]
where
[tex]a=2.5\ J/m^4\\\\b=12\ J/m^2[/tex]
As we know that
[tex]F=\dfrac{-dU}{dx}=-(4ax^3+2bx)[/tex]
Here, x= 0.5 m
So, it becomes,
[tex]F=-(4\times 2.5\times (0.5)^3+2\times 12\times 0.5)=-13.25\ N[/tex]
Now, we need to find the velocity when it crosses the origin,
[tex]F=mg\\\\F=m\dfrac{dv}{dt}\\\\F=mv\dfrac{dv}{dx}\\\\-(4ax^3+2bx)dx=mvdv\\\\\int\limits^{0.5}_0 {-(4ax^3+2bx)} \, dx =\int\limits {mv} \, dv\\\\ \dfrac{-4ax^4}{4}+\dfrac{2bx^2}{2}|_0^{0.5}=\dfrac{mv^2}{2}\\\\ax^4+bx^2|_0^{0.5}=\dfrac{mv^2}{2}\\\\2.5(0.5)^4+12(0.5)^2=\dfrac{2v^2}{2}\\\\0.15625+3=v^2\\\\3.15625=v^2\\\\\sqrt{3.15625}=v\\\\v=1.78[/tex]
Hence, a) -13.25 N, b) 1.78 m/sec
