Final answer:
None of the provided options are correct. To move a 110kg box with a coefficient of friction of 0.61, a force greater than 658.251 N is required, which exceeds all the given choices.
Step-by-step explanation:
To calculate the force required to push a 110kg box across the floor given a coefficient of friction of 0.61, we must identify the force of friction that needs to be overcome. This is given by the equation f = μ × N, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the box, which is mass (m) × gravity (g), so N = 110 kg × 9.81 m/s². The force of friction is therefore f = 0.61 × (110 kg × 9.81 m/s²). Solving this gives the minimum force that must be exceeded to move the box.
To find the minimum force required to push the box (ignoring acceleration), we calculate:
f = μ × N
f = 0.61 × (110 kg × 9.81 m/s²)
f = 0.61 × 1079.1 N
f = 658.251 N
Therefore, the force required must be greater than 658.251 N, which means none of the provided options (100 N, 200 N, 300 N, 400 N) are sufficient to push the box.