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. 261 b. 6 c. 24J d. 0J 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 />$W = F \cdot d = (5N)i + (4N)j \cdot (5m)i + (4m)j = 25Nm + 16Nm = 41Nm$<br /><br />The work done on the box increases its kinetic energy by the same amount:<br /><br />$KE_{final} = KE_{initial} + W = 15J + 41J = 56J$<br /><br />Therefore, the kinetic energy of the box at the end of displacement d is 56J.