Question

As shown in the diagram, the 15-kg box B moves to the right at an initial velocity vo = 2 m/s when d = 0.7 m. A force F = 120 N along a specified direction is applied on the box during the entire motion of the box along a smooth floor. When the box reaches the spring at A, it starts k to compress the spring until it comes to a temporary stop. d The spring constant is k = 1200 N/m. Solve the following: 30° a) Draw free body diagrams of the box before and after it reaches the spring. (( B b) Explicitly specify the condition for the spring to reach A the maximum compression. c) Use work and energy methods (either the principle of work and energy or conservation of energy, whichever fits the best) to determine the maximum compression of the spring.

Answer 1

Maximum compression occurs when the kinetic energy of the box is completely converted into potential energy of the spring.

= 15 lag, o=2m/s, d= 0.7m В F = 120 N d → F 30 amu A k= 1200 Ym (a) FTS fla F/2 4N x is the compression 4 ka Before After MBg mBg (6) For the spring to reach marimum compression, should be whole kinetic energy of box a converted into potential energy of spring. Work done by force F + Initial Kinetic Potential energy of spring at marimum compression. ( Work and Ff3 (d+x) + 1 B vo = 1 / 2 * x² Energy Principle) energy goon - 103.922 - 72.744 30 =0 600 x - - 103.920 102.744 = 0 x= 0.5093 m