The cart shown below moves across the table top as the block falls. What is the acceleration of the cart? Neglect friction and assume the following data:m1 = 2.2 kg,m2 = 3.7 kg,I = 0.5 kg · m2,andr = 19 cm.(Enter the magnitude in m/s2.) The image depicts a physics setup for a classic mechanics problem involving a pulley system.### Description:- **Components:** - A table with an object (a cart) labeled \( m_2 \) placed on top of it. - A pulley system attached to the edge of the table. - A mass \( m_1 \) hanging off the side of the table, connected by a string or rope to the cart \( m_2 \). - The string passes over the pulley.- **Pulley Details:** - The pulley is denoted to have specific properties: a moment of inertia (I) and radius (r).### Explanation:- **Setup:** - The cart \( m_2 \) is positioned on the table and connected to the hanging mass \( m_1 \) by a string. - The string runs over the pulley, which is depicted at the corner of the table. - **Mechanics:** - This system is typically used to study Newton's laws of motion, tension, and rotational dynamics. - The mass \( m_1 \) will exert a gravitational force causing the cart \( m_2 \) to accelerate along the table. - The pulley with moment of inertia \( I \) and radius \( r \) introduces rotational dynamics into the problem, affecting how the system accelerates.This setup is commonly explored in introductory physics courses to demonstrate concepts such as forces, tension in strings, acceleration, and rotational motion.

Name: The cart shown below moves across the table top as the block falls. Wh
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: The cart shown below moves across the table top as the block falls. What is the acceleration of the cart? Neglect friction and assume the following data:m1 = 2.2 kg,m2 = 3.7 kg,I = 0.5 kg · m2,andr = 19 cm.(Enter the magnitude in m/s2.) The image dep

Given data The mass of the first block is m1 = 2.2 kg. The mass of the second block is m2 = 3.7 kg. The moment of inertia of the pulley is I = 0.5 kgm2 The radius of pulley is r = 19 cm = 0.19 m The free-body diagram of the second block is shown below: The force equation in the horizontal direction is given as follows: T2=m2a The free-body diagram for the first block is given as: The force equation in the vertical direction is given as follows: m1a=m1g-T1T1=m1g-m1aThe expression for the torque equation is given as: T1-T2×r=I×arm1g-m1a-m2a=I×ar2 Substitute the value in the above equation. 2.2 kg×9.81 m/s2-2.2 kg×a-3.7 kg×a=0.5 kgm2×a0.19 m×0.19 m21.582=13.85a+2.2a+3.7aa=21.58213.85+2.2+3.7a=1.092 m/s2 Thus, the acceleration of the cart is 1.092 m/s2.