The coefficient of static friction between a block and a horizontal floor is \(0.40,\) while the coefficient of kinetic friction is \(0.15 .\) The mass of the block is \(5.0 \mathrm{kg} .\) A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?

: To find the force required to start the motion, first we need to calculate the force of static friction. The formula for static friction is: $$ f_s = \mu_s F_N $$ where \(f_s\) is the force of static friction, \(\mu_s\) is the coefficient of static friction, and \(F_N\) is the normal force. Since the block is on a horizontal surface and there's no vertical acceleration, we know that the normal force is equal to the gravitational force acting on the block: $$ F_N = m g $$ where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)). Now we can calculate the force of static friction: $$ f_s = \mu_s m g $$ Plugging in the values, we get: $$ f_s = (0.40)(5.0 \mathrm{kg})(9.8 \mathrm{m/s^2}) $$ Calculating, we get the force of static friction as: $$ f_s = 19.6 \mathrm{N} $$ So, the value of the applied horizontal force at the instant that the block starts to slide is \(19.6 \mathrm{N}\).

: Now let's determine the net force on the block after it starts to slide. Since the block is now in motion, we should consider the force of kinetic friction. The formula for kinetic friction is: $$ f_k = \mu_k F_N $$ where \(f_k\) is the force of kinetic friction, \(\mu_k\) is the coefficient of kinetic friction, and \(F_N\) is the normal force. We know the normal force is equal to the gravitational force acting on the block, so we can calculate the force of kinetic friction: $$ f_k = \mu_k m g $$ Plugging in the values, we get: $$ f_k = (0.15)(5.0 \mathrm{kg})(9.8 \mathrm{m/s^2}) $$ Calculating, we get the force of kinetic friction as: $$ f_k = 7.35 \mathrm{N} $$ Since the applied force is greater than the force of kinetic friction, the block will accelerate in the direction of the applied force. The net force on the block is the difference between the applied force and the force of kinetic friction: $$ F_{net} = F_{applied} - f_k $$ The applied force at the instant the block starts to slide (which we found in part (a)) and the force of kinetic friction are as follows: $$ F_{net} = 19.6 \mathrm{N} - 7.35 \mathrm{N} $$ Calculating, we get the net force on the block as: $$ F_{net} = 12.25 \mathrm{N} $$ So, the net force on the block after it starts to slide is \(12.25 \mathrm{N}\).