Final answer:
The force with which the hand pushes a 6kg box at a constant speed with a coefficient of friction of 0.78 is approximately 45.8 N.
Step-by-step explanation:
To find the force with which the hand pushes a 6kg box moving at a constant speed to the left, we need to consider the frictional force which is the product of the coefficient of friction and the normal force. Since the box moves at a constant speed, the pushing force must be equal to the frictional force due to Newton's first law of motion.
The normal force is equal to the weight of the box because there are no vertical accelerations, which is the mass of the box times the acceleration due to gravity (6 kg × 9.8 m/s²). This gives a normal force of 58.8 N. Then, we multiply this by the coefficient of friction which is 0.78, resulting in a frictional force of 45.864 N. Therefore, the force with which the hand must push the box is 45.864 N, which can be approximated to 45.8 N if rounding to one decimal place.