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 15J at the beginning of displacement d, what is its kinetic energy at the end of displacement d? Select one: a. 26J b. 6 C. 24 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, which states that the work done on an object is equal to the change in its kinetic energy.<br /><br />The work done on the box is given by the dot product of the force and the displacement:<br /><br />$W = F \cdot d$<br /><br />Substituting the given values:<br /><br />$W = (5N)i + (4N)j \cdot (5m)i + (4m)j = 25J + 16J = 41J$<br /><br />The initial kinetic energy of the box is given as 15J. Therefore, the change in kinetic energy is:<br /><br />$\Delta KE = W = 41J - 15J = 26J$<br /><br />So, the kinetic energy of the box at the end of displacement d is:<br /><br />$KE_{final} = KE_{initial} + \Delta KE = 15J + 26J = 41J$<br /><br />Therefore, the correct answer is:<br /><br />a. 26J