Alarge box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of kinetic friction is 0.30.
a. what is the minimum horizontal force needed to get the box moving?
b. if you continue to push with that force, what will the acceleration of the box be?
asap. you.

A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight). 

F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N

a = (735.75 N)/(500kg)= 1.4715 m/s^2

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Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then