Final answer:
The size of the force of kinetic friction acting on the 50 kg crate is calculated by multiplying the coefficient of kinetic friction (0.4) by the normal force, which is the weight of the crate. This gives a force of friction of 196 N, rounded to 200 N.
Step-by-step explanation:
To calculate the force of friction acting on a 50 kg crate pushed across a warehouse floor, we first need to determine the normal force. Since the only forces acting vertically are the weight of the crate (downward) and the normal force (upward), which balance each other out, the normal force equals the weight of the crate. The normal force (N) is given by the weight of the crate, which is the mass (m) times the acceleration due to gravity (g), so N = mg. In our case, N = (50 kg)(9.80 m/s²) = 490 N.
The force of kinetic friction (μ_k * N) is the product of the coefficient of kinetic friction and the normal force. Given that the coefficient of kinetic friction (μ_k) is 0.4, the force of kinetic friction is f_k = μ_k * N = (0.4)(490 N), which equals 196 N. Therefore, the size of the force of friction acting on the crate is 196 N, which corresponds to option A: 200 N (since options are given in whole numbers, we round to the nearest whole number).