Final answer:
The frictional force exerted on the block when the applied force is 15 N and the coefficient of friction is 0.2 can be calculated by the product of the coefficient of kinetic friction and the normal force. It is found to be most nearly 4 N. The correct option is b.
Step-by-step explanation:
The question given involves calculating the frictional force exerted on a block sliding on a horizontal tabletop, given the mass of the block, applied force, and the coefficient of kinetic friction. To calculate the frictional force (Ff), we can use the formula:
Ff = μk × N
where μk is the coefficient of kinetic friction and N is the normal force. The normal force is equal in magnitude to the gravitational force acting on the block (its weight), so in this case:
N = m × g = 2 kg × 9.8 m/s² = 19.6 N
With μk = 0.2:
Ff = 0.2 × 19.6 N = 3.92 N
Since we are asked to find which value it is most nearly, 4 N would be the closest approximation to the frictional force (Ff = 3.92 N). Option b. is the correct one.