91Ó°ÊÓ

The mass of an object is m=20 kg

The force applied to an object is$F=-3.0x-5.0x^2$

The potential energy of the object at x=0 is,$U_0=0J$

The velocity of an object is v=4.0 m/s in the negative direction of the x-axis when it is at x=0

Using the formula for work done on a system by an external force, we can find the stopping distance of an automobile.

Formulae:

The potential energy or work done by the system,$U\left(x\right)=∫F\left(x\right)dx$ (1)

The total energy of the system,$E=\frac{1}{2}m{v}^2+U\left(x\right)$ (2)

From the law of conservation of energy,

$E_i=E_f$ (3)

We are given that the force applied to an object is$F=-3.0x-5.0x^2$

Then P.E associated with it is using equation (1) is given as:

$$\begin{gathered} U\left(x\right)={∫}_{xi}^{xf}-3.0x-5.0x^{2}dx \\ =\left(-\frac{3.0x^2}{2}-\frac{5.0x^3}{3}\right)-U_0 \\ =\left(\frac{3.0x^2}{2}+\frac{5.0x^3}{3}\right)+U_0 \end{gathered}$$

$$\begin{gathered} \left|U\left(x=2.0m\right)\right|=\left(\frac{3.0{\left(2.0m\right)}^2}{2}+\frac{5.0{\left(2.0m\right)}^3}{3}\right)+0 \\ =19.4\mathrm{J} \end{gathered}$$

Therefore, the P.E. of the system associated with the force when the object is at x=2.0 m is equal to 19.4J

But, it has the speed of$v=4.0 \frac{m}{s}\mathrm{at}x=5.0 m$

From equation (3), we can get the initial speed of the object as follows:

$$\begin{gathered} E_i(x=0 m)=E_f(x=5 m) \\ \frac{1}{2}m{v}_0^2+U_0=\frac{1}{2}m{v}^2+U(x=5 m) \\ \frac{1}{2}\left(20kg\right)v_0^2+0\frac{1}{2}\left(20kg\right){\left(4.0\frac{m}{s}\right)}^2+\left(\frac{3.0{\left(5m\right)}^2}{2}+\frac{5.0{\left(5m\right)}^3}{3}+0\right) \\ {v}_0^2=40.58\frac{m^{2^2}}{s} \\ {v}_0=6.37\frac{m}{s} \end{gathered}$$

Therefore, the speed of an object when it passes through the origin is 6.37 m/s

From part a), we can write that,

$\left|U\left(x=2m\right)\right|=\left(\frac{3.0{\left(2m\right)}^2}{2}+\frac{5.0{\left(2m\right)}^3}{3}\right)+U_0$

If$U_0=-8J$then, the new potential energy is given as:

$$\begin{gathered} \left|U\left(x=2m\right)\right|=\left(\frac{3.0{\left(2m\right)}^2}{2}+\frac{5.0{\left(2m\right)}^3}{3}\right)-8J \\ =11.4J \end{gathered}$$

$v_0$will be the same as before because$U_0$on both sides of the equation part (b) cancels out.

So it is independent of$U_0$.

Therefore, the answer for part a) will be 11.4 J and that for part b) will be 6.37 m/s.