91Ó°ÊÓ

1. Graph of P.E (U) of the particle versus x and $\mathrm{U}\left(\mathrm{x}\right)={\mathrm{bx}}^2$
2. Mass of the particle, M = 2.0 Kg
3. The velocity of a particle at equilibrium${\mathrm{v}}_{\mathrm{m}}=85\mathrm{cm}/\mathrm{s}\mathrm{or}0.85\mathrm{m}/\mathrm{s}$.

Using the law of conservation of energy we can find the total mechanical energy of the particle from its maximum K.E energy. Next, from the graph, we can find the P.E (U) at some position. Applying the given relation to U, we can determine whether the particle will turn back before it reaches x=15 or not.

Formula:

The kinetic energy of a system$\mathrm{K}\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2.....\left(1\right)$

The law of conservation of energy gives E = constant .....(2)

From the equation, we can say that the total mechanical energy of the system is equal to the maximum K.E energy of the system, hence from equation (i), we get the total energy of the particle as:

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{{\mathrm{mv}}_{\mathrm{m}}}^2 \\ =\frac{1}{2}(2.0){(0.85)}^2 \\ =0.7225\mathrm{J} \end{gathered}$$

From the graph we can note that the P.E of the particle at x = 10 is 0.5 J.

i.e.$\mathrm{U}(\mathrm{x}=10)=0.5\mathrm{J}$

Since,

$$\begin{gathered} \mathrm{U}\left(\mathrm{x}\right)={\mathrm{bx}}^2 \\ \mathrm{b}=\frac{0.5}{{\left(10\right)}^2}9\mathrm{At}\mathrm{x}=10\mathrm{cm}) \\ =0.005\mathrm{J}/{\mathrm{cm}}^2 \\ \mathrm{Hence}, \\ \mathrm{U}(\mathrm{x})=\mathrm{E} \\ (0.005){\mathrm{x}}^2=0.72\mathrm{J} \\ \mathrm{x}=12\mathrm{cm} \end{gathered}$$

Therefore, the particle will not turn back before it reaches to 15 cm.

From step 3, it can be seen that the particle will be turn back at x= 12 cm.