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

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

Transcript text: Two carts with masses $m_{1}=100 \mathrm{~kg}$ and $\mathrm{m}_{2}=50 \mathrm{~kg}$ are connected by the rope at the applied force of $\mathrm{F}_{\mathrm{a}}=850 \mathrm{~N}$. Both cars are experiencing friction with coeffic of friction is 0.15

Solution

Solution Steps

Step 1: Identify the Forces Acting on the System

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.

Step 2: Calculate the Frictional Forces

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} $: \[ f_2 = \mu \cdot m_2 \cdot g = 0.15 \cdot 50 \cdot 9.81 = 73.575 \, \text{N} \]

Step 3: Calculate the Net Force and Acceleration

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: \[ a = \frac{F_{\text{net}}}{m_1 + m_2} = \frac{629.275}{150} = 4.1952 \, \text{m/s}^2 \]