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[SkyPhantom]

Refer to the given system, block b accelerate at 1m/s^2 the masses of block a and block b are 20 and 15 kg respectively find the frictional exerted on block A assuming pulleys to be frictionless and weightless

 

[🤖]

To find the frictional force exerted on block A, we need to first analyze the forces acting on both block A and block B.

For Block A:

• The force acting downwards is the weight of block A: \( F_{\text{weight A}} = m_A \cdot g \), where \( m_A = 20 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
• The force acting upwards is the tension in the rope: \( T \).
• The force of friction acting on block A is \( f_{\text{friction A}} \).

For Block B:

• The force acting downwards is the weight of block B: \( F_{\text{weight B}} = m_B \cdot g \), where \( m_B = 15 \, \text{kg} \).
• The force acting upwards is the tension in the rope: \( T \).
• The net force accelerating block B is \( F_{\text{net B}} = m_B \cdot a \), where \( a = 1 \, \text{m/s}^2 \).

Because the pulleys are frictionless and weightless, the tension in the rope is the same throughout. Therefore, we can set up equations of motion for both blocks using Newton's second law:

For Block A:
\[ T - f_{\text{friction A}} = m_A \cdot a \]

For Block B:
\[ T = m_B \cdot a + F_{\text{weight B}} \]

Now, we can substitute the values into these equations:
For Block A:
\[ T - f_{\text{friction A}} = 20 \cdot 1 \]
\[ T - f_{\text{friction A}} = 20 \]

For Block B:
\[ T = 15 \cdot 1 + 15 \cdot 9.81 \]
\[ T = 15 + 147.15 \]
\[ T = 162.15 \]

Now, substitute the value of T back into the equation for Block A:
\[ 162.15 - f_{\text{friction A}} = 20 \]
\[ f_{\text{friction A}} = 162.15 - 20 \]
\[ f_{\text{friction A}} = 142.15 \, \text{N} \]

Therefore, the frictional force exerted on block A is 142.15 N.