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
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Solution

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Solution Steps

Step 1: Identify the Forces Acting on the System

The two carts are connected and pulled by an applied force Fa=850N F_a = 850 \, \text{N} . Both carts experience friction, which opposes the motion. The frictional force f f for each cart can be calculated using the formula: f=μmg f = \mu \cdot m \cdot g where μ=0.15 \mu = 0.15 is the coefficient of friction, and g=9.81m/s2 g = 9.81 \, \text{m/s}^2 is the acceleration due to gravity.

Step 2: Calculate the Frictional Forces

For cart 1 with mass m1=100kg m_1 = 100 \, \text{kg} : f1=μm1g=0.151009.81=147.15N f_1 = \mu \cdot m_1 \cdot g = 0.15 \cdot 100 \cdot 9.81 = 147.15 \, \text{N}

For cart 2 with mass m2=50kg m_2 = 50 \, \text{kg} : f2=μm2g=0.15509.81=73.575N 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 Fnet F_{\text{net}} acting on the system is the applied force minus the total frictional force: Fnet=Fa(f1+f2)=850(147.15+73.575)=629.275N 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 m1+m2=100+50=150kg m_1 + m_2 = 100 + 50 = 150 \, \text{kg} .

The acceleration a a of the system can be calculated using Newton's second law: a=Fnetm1+m2=629.275150=4.1952m/s2 a = \frac{F_{\text{net}}}{m_1 + m_2} = \frac{629.275}{150} = 4.1952 \, \text{m/s}^2

Final Answer

The acceleration of the system is 4.1952m/s2\boxed{4.1952 \, \text{m/s}^2}.

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