A person pushes a box at a constant velocity across a floor: [](https://lh3.googleusercontent.com/fNDFJRNh_i73nD_EDcA4vHpEk_zNTi9DPht4aVohmb-05u6ZwZ2Hlquj6f8Miy11FqiiEwDsz6HTU-PQ_PUgtJPFczDBeJ0SRnQAE22I022H1LUw3PynWQ2zQv6gBUJJMiNZ2x2SQKoo1pOC5w) The box has a mass of 40 kg, and the coefficient of kinetic friction between the box and the floor is 0.35. What is the magnitude of the force that the person exerts on the box? ### ==Answer== The box is moving at a constant velocity, which means there is no acceleration, and therefore no net force on the box. This means the force exerted by the person is exactly equal to the force of friction. The force of friction between the box and the floor is given by the equation: $\vec{F}_f = \mu{_k} \cdot \vec{F}_N$ The normal force is equal in magnitude to the weight of the box ( $\vec{F}_g$ ), which is given by the equation: $\vec{F}_N$ = $\vec{F}_g$ = _ma_ =(40 kg) $\cdot$ (10 $\frac{m}{s^2}$ )= 400 N Therefore, the force of friction is: $\vec{F}_f = \mu{_k} \cdot \vec{F}_N$ $\vec{F}_f$= (0_._35) $\cdot$ (400 N) = 140 N