To find the force exerted by block A on block B, we'll start by analyzing the forces acting on both blocks. Since the surface is frictionless, the only horizontal force acting on block A and block B is the applied force of 10 N.
First, calculate the total mass of the system:
\(m_A = 2 \text{ kg}\)
\(m_B = 3 \text{ kg}\)
Total mass, \(M = m_A + m_B = 2 \text{ kg} + 3 \text{ kg} = 5 \text{ kg}\)
Next, find the acceleration of the system using Newton's second law:
\(F = M \cdot a\)
\(10 \text{ N} = 5 \text{ kg} \cdot a\)
Solving for \(a\) gives:
\(a = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m/s}^2\)
Now, focus on block B. The only force responsible for the acceleration of block B is the force exerted by block A on it.
Use Newton's second law for block B:
\(F_{AB} = m_B \cdot a\)
\(F_{AB} = 3 \text{ kg} \cdot 2 \text{ m/s}^2 = 6 \text{ N}\)
Thus, the force exerted by block A on block B is \(6 \, \text{N}\).
Step 1: Calculate the acceleration of the system.
- The total mass of the system is: M = mA + mB = 2kg + 3kg = 5kg. - Using Newton’s second law: \[ F = M \cdot a \Rightarrow a = \frac{F}{M} = \frac{10}{5} = 2 \text{ m/s}^2 \] Step 2: Calculate the force exerted by block A on block B. - The force on block B due to block A is: \[ F_B = m_B \cdot a = 3 \times 2 = 6N \] Step 3: Conclude. - The force exerted by block A on block B is 6N.
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: