A box is sliding on the frictionless floor through a displacement d=(5m)i+(4m)j while a constant force F=(5N)i+(4N)j acts on the box. If the box has kinetic energy 15 at the beginning of displacement d, what is its kinetic energy at the end of displacement d? Select one: a. 26 b. 6J C. 24J d. 0J e. 56J

To find the kinetic energy of the box at the end of displacement d, we can use the work-energy principle. The work done by the force F on the box is equal to the change in kinetic energy of the box.<br /><br />The work done by the force F is given by the dot product of the force and the displacement:<br /><br />Work = F · d = (5N)i + (4N)j · [(5m)i + (4m)j] = 5 * 5 + 4 * 4 = 25 + 16 = 41 Joules<br /><br />The initial kinetic energy of the box is given as 15 Joules.<br /><br />Therefore, the kinetic energy at the end of displacement d is:<br /><br />Final kinetic energy = Initial kinetic energy + Work done = 15J + 41J = 56J<br /><br />So, the correct answer is:<br /><br />e. 56J