Figure shows a plot of potential energy Uversus position xof ag particle that can travel only along an xaxis. (Nonconservative forces are not involved.) Three values are, UA= 15.0J, UB = 35.0 Jand UC = 45.0 J. The particle is released at x= 4.5 mwith an initial speed of, headed in the negative xdirection. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive xdirection when it is released at x = 4.5 mat speed 7.0 m/s. (d) If the particle can reach x= 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of?

$$\begin{gathered} 1.Themassofparticleis,m=0.90kg \\ 2.Thespeedofaparticleatx=4.5mis,v=7.0m/s \\ 3.U_A=15.00\mathrm{J} \\ 4.U_B=35.00\mathrm{J} \\ 5.U_C=45.00\mathrm{J} \end{gathered}$$

Use the principle of conservation of energy to find the required parameters.According to the law of energy conservation, energy can neither be created, nor be destroyed. Calculate the kinetic energy at any point along the path of the particle using conservation of energy. Then, using the equation of kinetic energy, calculate the speed at the required position. To find the position of the turning point and force, consider the slope of the line having the point of potential energy greater than total energy and less than total energy.

Formulae are as follow:

$$\begin{gathered} U_1+K_1=U_2+K_2 \\ K=\frac{1}{2}m{v}^2 \\ F=-\frac{∆y}{∆x} \end{gathered}$$

Where, K is kinetic energy, Uis potential energy, m is mass, v is velocity, g is acceleration due to gravity $∆y,∆x$and are change in displacements.

The speed of the particle at x = 4.5 m is given as v = 7.0 m/s. So, the kinetic energy of a particle at this position is,

$$\begin{gathered} K=\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.90\times 7.0^2 \\ k=22.05\mathrm{J} \end{gathered}$$

From the graph that the potential energy of a particle at point x= 4.5 is,

$U_A=15.00\mathrm{J}$

So, the total energy of a particle is,

$$\begin{gathered} E=K+U=22.05+15.00 \\ E=37.05\mathrm{J} \end{gathered}$$

As the potential energy at point x = 1.0m is less than the total mechanical energy (E = 37.05 J), the particle can reach that point.

And its kinetic energy will be,

$K=E-U_B=2.05\mathrm{J}$

So, the speed of a particle is,

$$\begin{gathered} v=\sqrt{\frac{2\times K}{m}}=\sqrt{\frac{2\times 2.05}{0.90}} \\ v=2.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the particle can reach x =1.0 m, and its speed is v = 2.1 m/s

Now,
$F=\frac{∆U}{∆x}$

As the graph is U vs x, this is the negative of slope.

So, let us consider the motion of particle to the left of point x = 4.5 m

$$\begin{gathered} F=-\frac{35.00-15.00}{2.0-4.0} \\ F=10\mathrm{N} \end{gathered}$$

Hence,the magnitude of the force on the particle as it begins to move to
the left of x= 4.0 m, is F = 10N .

As the force acting on the particle is positive, the direction of force is along positive x axis.

The potential energy at point x = 7m ($U_C=45.00\mathrm{J}$) is greater than the total energy E = 37.05 J.

So, the particle cannot reach there.

So, to calculate the turning point, consider the motion of particle from x = 5 m to x = 6 m.

Considering the slope of a line AB,

$$\begin{gathered} \frac{45.0-15.0}{6.0-5.0}=\frac{37.05-15.0}{x-5.0} \\ \mathrm{Solving}\mathrm{this}\mathrm{equation}\mathrm{for}x, \\ x=5.7\mathrm{m} \end{gathered}$$

Hence, the particle cannot reach x= 7.0 m. The turning point is x = 5.7m

In a similar way to part b, consider the slope of graph.

So, considering the slope of line AB in above figure,

$F=-\frac{45.0-15.0}{6.0-5.0}$

F = -30.0 N

So, the magnitude of force is, $\left|F\right|=30.0\mathrm{N}$.

As the force calculated in part e is negative, the direction of force is along the negative x axis.

Therefore, the speed of a particle at any position can be found using the equation of conservation of energy. To calculate the turning point and force,the slope of line can be used.