Let’s analyze the forces acting on the blocks:
Calculate the Total Mass: The total mass of the system \( m \):
\[ m = m_A + m_B + m_C = 5 \, \text{kg} + 3 \, \text{kg} + 2 \, \text{kg} = 10 \, \text{kg}. \]Calculate the Acceleration of the System: Using Newton’s second law \( F = ma \):
\[ a = \frac{F}{m} = \frac{80 \, \text{N}}{10 \, \text{kg}} = 8 \, \text{m/s}^2. \]Calculate the Tension \( T_2 \) in the String Connecting B and C: For block C (mass = 2 kg), using
\[ F = ma: \] \[ T_2 = m_C \times a = 2 \, \text{kg} \times 8 \, \text{m/s}^2 = 16 \, \text{N}. \]Calculate the Tension \( T_1 \) in the String Connecting A and B: The force acting on block B (mass = 3 kg) includes both its weight and the tension \( T_2 \):
\[ T_1 = m_B \times a + T_2 = (3 \, \text{kg} \times 8 \, \text{m/s}^2) + 16 \, \text{N} = 24 \, \text{N} + 16 \, \text{N} = 40 \, \text{N}. \]Therefore, for block A (mass = 5 kg):
\[ T_1 = m_A \times a + T_1 + T_2 = (5 \, \text{kg} \times 8 \, \text{m/s}^2) = 40 \, \text{N} + T_1. \]Calculate Final Tensions: Now, substituting for \( T_2 \):
\[ T_1 = 5 \times 8 \, \text{N}, T_2 = 40 + 8 \times 3 = 64 \, \text{N}. \]The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.
The least acidic compound, among the following is
Choose the correct set of reagents for the following conversion:
The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.
The formula for the Centre of Mass:
The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.