2. A particle of mass 2. Okg moves along the x axis through a region in which its potential energy U(x) varies as shown in Fig. 6. When the particle is at -5F-=-- x=2.0m, its velocity is -20m/s (a)Calculate the force acting on the particle at this position. (b)Between.10 what limits does the motion take place?(c)How fast is s it moving when it is at x7. 0m? Solution (a) Using the relationship of F =-,(PE),from Fig6 the diagram, the force acting on the particle at x=2.0m the opposite of the tangent of the line L12. So F (PB)≈17-3 4.7N (b) At the point x20m, we can calculate the total mechanical energy E=PE+Ke=-8 As we must have E> PE, PE s-4J Look at the diagram, the position limit of the particle is 1.5m≤x≤141 (c)E=PE+KE→-17+×2xy2=-4→w=√13(m/s) 3. A projectile is launched at initial speed vo at an angle 0 with the horizontal direction. Neglect air friction and assume the trajectory is confined to a region near the surface of the Earth. Find the maximum vertical height of the projectile at the top of its flight path by using CWe theorem Solution Set up the coordinate system shown in figure, choose the elevation of the y -coordinate origin The initial kinetic energy KE=mvo The initial potential energy is PE=mgy=0 The kinetic energy at the top of its flight path is KEop=m(vo cos 0) The potential energy at the top of its flight path is PErop=mgy op Making the appropriate substitution into the Cwe theorem2. A particle of mass 2.0kg moves along the x axis through a region in which its potential energy U(x) varies as shown in Fig. 6. When the particle is at x=2.0m, its velocity is -2.0m/s. (a) Calculate the force acting on the particle at this position. (b) Between what limits does the motion take place? (c) How fast is it moving when it is at x=7.0m? Solution: (a) Using the relationship of ( ) d d PE x Fx = − , from the diagram, the force acting on the particle at x=2.0m is the opposite of the tangent of the line L12. So 4.7N 3 17 3 ( ) d d ≈ − = − PE = x Fx (b) At the point x=2.0m, we can calculate the total mechanical energy: 2 2 4 (J) 2 1 8 2 E = PE + KE = − + × × = − As we must have E ≥ PE , PE ≤ −4J .Look at the diagram , the position limit of the particle is 1.5m ≤ x ≤ 14m . (c) 2 4 13 (m/s) 2 1 17 2 E = PE + KE ⇒ − + × × v = − ⇒ v = 3. A projectile is launched at initial speed 0 v at an angle θ with the horizontal direction. Neglect air friction and assume the trajectory is confined to a region near the surface of the Earth. Find the maximum vertical height of the projectile at the top of its flight path by using CWE theorem. Solution: Set up the coordinate system shown in figure, choose the elevation of the y –coordinate origin. The initial kinetic energy is 2 0 2 1 KE mv i = The initial potential energy is PEi = mgyi = 0 The kinetic energy at the top of its flight path is 2 0 ( cos ) 2 1 KEtop = m v θ The potential energy at the top of its flight path is PEtop = mgytop Making the appropriate substitution into the CWE theorem, y x Fig.6 0 5 10 15 -20 -15 -10 -5 0 U(x) (J) x (m) 1 2