Question

22: A 2.0 kg object moves along the x-axis under the influence of an unknown force. Its potential energy due to the force is U(x) = ax4 + bx2 , where a = 2.5 J/m4 and b = 12 J/m2 a) What is the force on the object when x = 0.5 m? b) If the object is released from rest at x = 0.5 m, what is its velocity when it crosses the origin?

Answer 1

Answer: a) -13.25 N, b) 1.78 m/sec

Step-by-step explanation:

Since we have given that

$U(x)=ax^4+bx^2$

where

$a=2.5\ J/m^4\\\\b=12\ J/m^2$

As we know that

$F=\dfrac{-dU}{dx}=-(4ax^3+2bx)$

Here, x= 0.5 m

So, it becomes,

$F=-(4\times 2.5\times (0.5)^3+2\times 12\times 0.5)=-13.25\ N$

Now, we need to find the velocity when it crosses the origin,

$F=mg\\\\F=m\dfrac{dv}{dt}\\\\F=mv\dfrac{dv}{dx}\\\\-(4ax^3+2bx)dx=mvdv\\\\\int\limits^{0.5}_0 {-(4ax^3+2bx)} \, dx =\int\limits {mv} \, dv\\\\ \dfrac{-4ax^4}{4}+\dfrac{2bx^2}{2}|_0^{0.5}=\dfrac{mv^2}{2}\\\\ax^4+bx^2|_0^{0.5}=\dfrac{mv^2}{2}\\\\2.5(0.5)^4+12(0.5)^2=\dfrac{2v^2}{2}\\\\0.15625+3=v^2\\\\3.15625=v^2\\\\\sqrt{3.15625}=v\\\\v=1.78$

Hence, a) -13.25 N, b) 1.78 m/sec