A 4kg box is pulled by 20N force along a flat table. The box is in limiting equilibrium. What is the coefficient of friction?

Steps to Solve

Here's how we solve question 1:

1. Identify the knowns We have mass $m = 6 \text{ kg}$, coefficient of kinetic friction $\mu_k = 0.4$, and acceleration $a = 2 \text{ m/s}^2$.

2. Calculate the friction force The friction force is given by $F_f = \mu_k \times N$, where $N$ is the normal force. Since the surface is horizontal, the normal force equals the weight of the object, $N = mg$. Therefore, $F_f = \mu_k \times mg = 0.4 \times 6 \text{ kg} \times 9.8 \text{ m/s}^2 = 23.52 \text{ N}$.

3. Apply Newton's second law Newton's second law states that $F_{net} = ma$, where $F_{net}$ is the net force acting on the object. The net force is the applied force $F_a$ minus the friction force $F_f$, so $F_{net} = F_a - F_f$. Thus, $F_a - F_f = ma$.

4. Solve for the applied force We want to find $F_a$, so we rearrange the equation to get $F_a = ma + F_f = (6 \text{ kg} \times 2 \text{ m/s}^2) + 23.52 \text{ N} = 12 \text{ N} + 23.52 \text{ N} = 35.52 \text{ N}$.