Suppose you have a 144-kg wooden crate resting on a wood floor. (For each answer, enter a number. $\mu_{k}=0.3$ and $\mu_{s}=0.5$ )
705.6N is the maximum force you can exert horizontally on the crate without moving it.
(b) If you continue to exert this force (in $\mathrm{m} / \mathrm{s}^{2}$ ) once the crate slarts to slip, what will the magnitude of its acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) then be? $\mathrm{m} / \mathrm{s}^{2}$

Step 1 ：Given that the mass of the crate (m) is 144 kg, the acceleration due to gravity (g) is 9.8 m/s², the coefficient of static friction (μs) is 0.5, the coefficient of kinetic friction (μk) is 0.3, and the applied force (Fapplied) is 705.6 N.

Step 2 ：First, calculate the normal force (FN), which is equal to the weight of the crate. This can be found using the equation \(F_N = m \cdot g\), which gives \(F_N = 144 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 1411.2 \, \text{N}\).

Step 3 ：The force of static friction (fs), which keeps the crate from moving, can be calculated using the equation \(f_s = \mu_s \cdot F_N\). Substituting the given values gives \(f_s = 0.5 \cdot 1411.2 \, \text{N} = 705.6 \, \text{N}\).

Step 4 ：Once the crate starts to move, the force of kinetic friction (fk) comes into play. This can be calculated using the equation \(f_k = \mu_k \cdot F_N\). Substituting the given values gives \(f_k = 0.3 \cdot 1411.2 \, \text{N} = 423.36 \, \text{N}\).

Step 5 ：The net force acting on the crate (Fnet) is the applied force minus the force of kinetic friction. This gives \(F_{net} = F_{applied} - f_k = 705.6 \, \text{N} - 423.36 \, \text{N} = 282.24 \, \text{N}\).

Step 6 ：Finally, the magnitude of the crate's acceleration (a) once it starts to slip can be found using Newton's second law, \(a = \frac{F_{net}}{m}\). Substituting the given values gives \(a = \frac{282.24 \, \text{N}}{144 \, \text{kg}} = 1.96 \, \text{m/s}^2\).

Step 7 ：\(\boxed{1.96 \, \text{m/s}^2}\) is the magnitude of the crate's acceleration once it starts to slip.