A force parallel to the x-axis acts on a particle movingalong the x-axis. This force produces potential energy U(x) given byU(x) = ax4, where a = 0.630 J/m4. What is the force (magnitude anddirection) when the particle is at x = -0.800 m?
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A force parallel to the x-axis acts on a particle moving
along the x-axis. This force produces potential energy U(x) given by
U(x) = ax4, where a = 0.630 J/m4. What is the force (magnitude and
direction) when the particle is at x = -0.800 m?
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- A pogo stick has a spring with a spring constant of 2.5 × 104 N/m, which can be compressed 12.0 cm. To what maximum height from the uncompressed spring can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40 kg?The figure here shows a plot of potential energy U versus position x of a 0.876 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, Ug = 35.0 J and Uc = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.96 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 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 = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.96 m/s. (d) If the particle can reach x = 7.00 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 x = 5.00 m? UcT UB 4 x (m) (a) Number i Unit (b) Number i Unit (c) (d) Number i…A 0.800-kg particle has a speed of 2.10 m/s at point and kinetic energy of 7.80 J at point . (a) What is its kinetic energy at ?(b) What is its speed at ?(c) What is the net work done on the particle by external forces as it moves from to ?
- A 7.00-kg particle moves from the origin to position ©, having coordinates x = 6.50 m and y = 6.50 m as shown in the figure. One force on the particle is the gravitational force acting in the negative y direction. Using the equation (W = FAr cos 0 = F · Ař), calculate the work done by the gravitational force in going from O to © along the following paths. у (m) B (х, у) x (m) (a) the purple path (o0©) (b) the red path (O®© (c) the blue path (O© (d) Your results should all be identical. Why? This answer has not been graded yet.The figure here shows a plot of potential energy U versus position x of a 0.898 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and Uc = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.52 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 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 = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.52 m/s. (d) If the particle can reach x = 7.00 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 x = 5.00 m? Uc UB UA 2 4 x (m) (a) Number i Unit (b) Number i Unit (c) (d)…Problem Thirteen. A potential energy function is given by U(r)=U2 where o = 2.00 m r r and U, = 1.00 J. Find the magnitude of the force at r= 1.00 m. Give an answer in kN. 25.) (A) 280 (В) 106 (С) 420 (D) 470 (E) 530
- The figure below shows a plot of potential energy U versus position x of a 0.96 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are UA = 10.0 J, Ug = 25.0 J, and UC = 35.0 J. (r) Ucr UB 2 4 6 x (m) The particle is released at x = 4.5 m with an initial speed of 8.5 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? ---Select-- C ---Select- (b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? magnitude N direction Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 8.5 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? Select CIf the potential energy of a particle is given by U (r, y, 2) = –kry?2³, find the force associated with this potential energy where k is a positive constant. k(2x²yz³i+y²z³j + xy²2²k) kry²2*(i + ĵ + k) k(y²2³i + 2xyz³j + 3xy²2²k) -kry 2*(i +j+ k) 33 3 k(y²2*i+ xyz°j +xy²2² k). A pogo stick has a spring with a spring constant of 2.5×104N/m,, which can be compressed 12.0 cm. To what maximum height from the uncompressed spring can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40 kg?
- The figure here shows a plot of potential energy U versus position x of a 0.878 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA-15.0J, Us - 35.0 J and Uc-45.0 J. The particle is released at x -4.50 m with an initial speed of 7.99 m/s, headed in the negative x direction. (a) If the particle can reach x -1.00 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- 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x - 4.50 m at speed 7.99 m/s. (d) If the particle can reach x - 7.00 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 x-5.00 m? (0)2 Ucr UB UA 2 (a) Number (b) Number 10 4 x (m) (c) positive (+x) (e)…The figure here shows a plot of potential energy U versus position x of a 0.877 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, Ug = 35.0 J and Uc = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.39 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 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 = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.39 m/s. (d) If the particle can reach x = 7.00 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 x = 5.00 m? Uc UB UA 4. x (m) (a) Number | Unit (b) Number Unit (c) (d) Number…The figure here shows a plot of potential energy U versus position x of a 0.898 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and Uc = 45.0 J. The particle is released atx = 4.50 m with an initial speed of 7.91 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 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 = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.91 m/s. (d) If the particle can reach x = 7.00 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 x = 5.00 m? (1) 2 Uc UB UA + 2 (a) Number (b) Number 4 x (m) 4.24 10 1 6 Unit…