A horizontal force 10 N is applied to a block 4 as shown in figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
The total mass of the system is: M = mA + mB = 2kg + 3kg = 5kg. - Using Newton’s second law:F=M⋅a⇒a=FM=105=2 m/s2F=M⋅a⇒a=MF=510=2 m/s2
Step 2:
Calculate the force exerted by block A on block B. - The force on block B due to block A is:FB=mB⋅a=3×2=6NFB=mB⋅a=3×2=6N
Step 3:
Conclude. - The force exerted by block A on block B is 6N.
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Approach Solution -2
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.
