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Q104P

Expert-verifiedFound in: Page 211

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A 20 kg**** 0object is acted on by a conservative force given by ${\mathbf{F}}{\mathbf{=}}{\mathbf{-}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{x}}{\mathbf{-}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{{\mathbf{x}}}^{{\mathbf{2}}}$****, with F**** in newton and x in meters. Take the potential energy associated with the force to be zero when the object is at x=0****. (a) What is the potential energy of the system associated with the force when the object is at x=2.0 m****? (b) If the object has a velocity of 4.0 m/s**** in the negative direction of the x-axis when it is at x=5.00 m****, what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be -8.0 J**** when the object is at x=0****?**

a) The P.E. of the system associated with the force when the object is at x=2.0 m is 19.4 J

b) The speed of the object when it passes through the origin is 6.37 m/s

c) Answer for part a) will be 11.4 J and that for part b) will be 6.37 m/s when P.E. of the system when the object is at x=0 is-8.0 J.

The mass of an object is m=20 kg

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

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,$U\left(x\right)=\int F\left(x\right)dx$ (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$F=-3.0x-5.0{x}^{2}$

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

$U\left(x\right)={\int}_{xi}^{xf}-3.0x-5.0{x}^{2}dx\phantom{\rule{0ex}{0ex}}=\left(-\frac{3.0{x}^{2}}{2}-\frac{5.0{x}^{3}}{3}\right)-{U}_{0}\phantom{\rule{0ex}{0ex}}=\left(\frac{3.0{x}^{2}}{2}+\frac{5.0{x}^{3}}{3}\right)+{U}_{0}$

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

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 $v=4.0\hspace{0.33em}\frac{m}{s}\mathrm{at}x=5.0\hspace{0.33em}m$

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

${E}_{i}(x=0\hspace{0.33em}m)={E}_{f}(x=5\hspace{0.33em}m)\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{0}^{2}+{U}_{0}=\frac{1}{2}m{v}^{2}+U(x=5\hspace{0.33em}m)\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(20kg\right){v}_{0}^{2}+0\frac{1}{2}\left(20kg\right){\left(4.0\frac{m}{s}\right)}^{2}+\left(\frac{3.0{\left(5m\right)}^{2}}{2}+\frac{5.0{\left(5m\right)}^{3}}{3}+0\right)\phantom{\rule{0ex}{0ex}}{v}_{0}^{2}=40.58\frac{{m}^{{2}^{2}}}{s}\phantom{\rule{0ex}{0ex}}{v}_{0}=6.37\frac{m}{s}$

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:

$\left|U\left(x=2m\right)\right|=\left(\frac{3.0{\left(2m\right)}^{2}}{2}+\frac{5.0{\left(2m\right)}^{3}}{3}\right)-8J\phantom{\rule{0ex}{0ex}}=11.4J$

${v}_{0}$ 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.

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