A single conservative force acts on a 4.70−kg particle within a system due to its interaction with the rest of the system. The equation FX = 2 x+4 describes the force, where FX is in newtons and x is in meters. As the particle moves along the x axis from x = 0.96 m to x = 5.70 m, calculate the following. (a) the work done by this force on the particle (b) the change in the potential energy of the system J (c) the kinetic energy the particle has at x = 5.70 m if its speed is 3.00 m/s at x = 0.96 m J
![A plot of potential energy versus position is shown for a 0.27 kg particle that can move along an x axis as it is acted upon by a conservative force. In the graph, UA = 9.9 J, UC = 20 J and UD = 31 J (corresponding to x > 8 ). The particle is observed at the point where the potential energy U forms a curved "hill" to have kinetic energy of 4.22 J (At this point UB = 14 J). What is the difference in particle speed when it is located at x = 6.5 m compared to x = 3.5 m ? Assume mechanical energy is conserved. [Find the speed for x = 6.5 m and subtract the speed for x = 3.5 m] Δv = m/s What is the range of possible particle motion? [Find the turnaround location for the left and right side. The particle turns around when v = 0. Take the difference in these (right value minus left value) to get the range of the particle's motion] |Δx| = m](https://www.doubtrix.com/uploads/editor/2476129399HnqfCEtbEQ.png)
![A plot of potential energy versus position is shown for a 0.203 kg particle that can move along an x-axis as it is acted upon by a conservative force. In the graph, UA = 8.6 J, UC = 23 J and UD = 30 J (corresponding to x > 8 ). The particle is observed at the point where the potential energy U forms a curved "hill" to have kinetic energy of 4.98 J (At this point UB = 13 J). What is the difference in particle speed when it is located at x = 6.5 m compared to x = 3.5 m ? Assume mechanical energy is conserved. [Find the speed for x = 6.5 m and subtract the speed for x = 3.5 m] Δv = m/s What is the range of possible particle motion? [Find the turnaround location for the left and right side. The particle turns around when v = 0. Take the difference in these (right value minus left value) to get the range of the particle's motion] |Δx| = m](https://www.doubtrix.com/uploads/editor/9561212003pxsDdCqdzA.png)