Questions: Two carts with masses m₁=100 kg and m₂=50 kg are connected by the rope at the applied force of Fₐ=850 N. Both cars are experiencing friction with coeffic of friction is 0.15

Solution Steps

The two carts are connected and pulled by an applied force $F_a = 850 \, \text{N}$. Both carts experience friction, which opposes the motion. The frictional force $f$ for each cart can be calculated using the formula: $$f = \mu \cdot m \cdot g$$ where $\mu = 0.15$ is the coefficient of friction, and $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.

For cart 1 with mass $m_1 = 100 \, \text{kg}$: $$f_1 = \mu \cdot m_1 \cdot g = 0.15 \cdot 100 \cdot 9.81 = 147.15 \, \text{N}$$

For cart 2 with mass $m_2 = 50 \, \text{kg}$: $$f_2 = \mu \cdot m_2 \cdot g = 0.15 \cdot 50 \cdot 9.81 = 73.575 \, \text{N}$$

The net force $F_{\text{net}}$ acting on the system is the applied force minus the total frictional force: $$F_{\text{net}} = F_a - (f_1 + f_2) = 850 - (147.15 + 73.575) = 629.275 \, \text{N}$$

The total mass of the system is $m_1 + m_2 = 100 + 50 = 150 \, \text{kg}$.

The acceleration $a$ of the system can be calculated using Newton's second law: $$a = \frac{F_{\text{net}}}{m_1 + m_2} = \frac{629.275}{150} = 4.1952 \, \text{m/s}^2$$

Final Answer