Final answer:
The force of friction when pushing a 50kg box with 15N of force and a coefficient of static friction of 0.5 is 15N, because it's less than the maximum static frictional force which would be 245N.
Step-by-step explanation:
If you are pushing a 50kg box with 15N of force, and the coefficient of static friction is 0.5, to find the force of friction, we first need to calculate the normal force. The normal force (N) is equal to the mass of the box (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s2. Hence, N = m * g = 50kg * 9.8 m/s2 = 490 N.
Now, the maximum static frictional force (fs) can be calculated using the coefficient of static friction (μs) by the equation fs = μs * N. Substituting the known values yields fs = 0.5 * 490 N = 245 N. This means the maximum force of static friction that can act is 245N. However, since only 15N of force is being applied, and it's less than the maximum possible static friction, the actual force of friction opposing the motion will be equal to the force applied, which is 15N.
Therefore, the correct answer is D) 15N.