Chapter 10 - Energy - Exercises and Problems - Page 275: 58

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a) We assume that the object is an isolated system, so its net energy is constant. We know that the particle is moving in the $x$-axis with a kinetic energy of 12 J. This means that on the horizontal level, its net energy is 12 J. Thus, the turning point occurs when its potential energy reaches 12 J which means that its kinetic energy is fully transferred to a potential energy. In the given graph, this turning happens at two points, at $$x=\color{red}{\bf 1}\;\rm m$$ and at $$x=\color{red}{\bf 8}\;\rm m$$ _________________________________________________ b) We know that the net energy of the system is constant, so $$E=K+U=12\;\rm J$$ At $x=6\;\rm m$, its potential energy is 0 J, so that $$K+0=\frac{1}{2}mv^2+0=12$$ Hence, $$v=\sqrt{\dfrac{2\times 12}{m}}=\sqrt{\dfrac{2\times 12}{0.5}}$$ $$v=\color{red}{\bf 6.93}\;\rm m/s$$ _________________________________________________ c) The particle's maximum speed occurs at the particle's maximum kinetic energy which occurs when the particle's potential energy is at its lowest value. Thus, the maximum speed of the particles occurs at $x=6\;\rm m$ at which its potential energy is zero. Thus, $$v_{max}=\sqrt{\dfrac{2\times 12}{m}}=\sqrt{\dfrac{2\times 12}{0.5}}$$ $$v_{max}=\color{red}{\bf 6.93}\;\rm m/s$$ _________________________________________________ d) First, we need to find a general velocity formula of our object relative to the given data and the given figure. We know that $$K+U=12$$ Hence, $$\frac{1}{2}mv^2+U=12$$ Thus, $$v=\sqrt{\dfrac{24-2U}{ m}}$$ At the end of the first meter, $x=1$, $U=2$, so $$v_{x=1}=\sqrt{\dfrac{24-2(2)}{ 0.5}}=\bf 6.3\;\rm m/s$$ And so on... $$v_{x=2}=\sqrt{\dfrac{24-2(4)}{ 0.5}}=\bf 5.66\;\rm m/s$$ $$v_{x=3}=\sqrt{\dfrac{24-2(5)}{ 0.5}}=\bf 5.3\;\rm m/s$$ $$v_{x=4}=\sqrt{\dfrac{24-2(8)}{ 0.5}}=\bf 4\;\rm m/s$$ $$v_{x=5}=\sqrt{\dfrac{24-2(4)}{ 0.5}}=\bf 5.66\;\rm m/s$$ $$v_{x=6}=\sqrt{\dfrac{24-2(0)}{ 0.5}}=\bf 6.93\;\rm m/s$$ $$v_{x=7}=\sqrt{\dfrac{24-2(5)}{ 0.5}}=\bf 5.3\;\rm m/s$$ $$v_{x=7}=\sqrt{\dfrac{24-2(12)}{ 0.5}}=\bf 0\;\rm m/s$$ The object slows down while moving to the right from $x=1$ m to $x=4$ m, then it speeds up from $x=5$ m to $x=6$ m, then it slows down again from $x=7$ m until it stops at $x=8$ m. _________________________________________________ e) If $E=4$ J only, it will move from $x=1$ m to $x=2$ m or from $x=5$ m to $x=7$ m; at which Its speed is zero at $x=2,5,7$.