Final answer:
To determine the force exerted by a woman pushing a chest at a constant speed, we calculate the force of kinetic friction. The force exerted by the woman is found to be 243 N, which matches option A as the correct answer.
Step-by-step explanation:
The question asks us to find the force exerted by a woman who pushes an oak chest across an oak floor at a constant speed. To solve for the force exerted by the woman, we must use the concept of kinetic friction and Newton's second law of motion. The force exerted by the woman must be equal to the force of kinetic friction, since the chest is moving at a constant speed, which implies that the net force acting on the chest is zero.
The force of kinetic friction (f_k) can be found using the equation f_k = μ_k × N, where μ_k is the coefficient of kinetic friction and N is the normal force. Since the chest is moving horizontally, the normal force (N) is equal to the weight of the chest (w = m × g), where m is the mass of the chest and g is the acceleration due to gravity (9.8 m/s² on Earth). Therefore, we can calculate the force as follows: f_k = μ_k × m × g.
By substituting the given values (μ_k = 0.620, m = 40.0 kg, g = 9.8 m/s²), we find the force of kinetic friction: f_k = 0.620 × 40.0 kg × 9.8 m/s² = 243.04 N. This result is very close to option A, which is 243 N.
Therefore, the force exerted by the woman is approximately 243 N, making option A the correct option.