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. 0) e. 56

The correct answer is:<br /><br />c. 24J<br /><br />Explanation:<br />The work done by the force F on the box 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 />Since the box starts with kinetic energy 15J and the work done on it is 41J, the total kinetic energy at the end of the displacement is:<br /><br />KE_final = KE_initial + Work = 15J + 41J = 56J<br /><br />Therefore, the correct answer is:<br /><br />e. 56