Questions: A force Fapp=25 N is applied to block A of mass mA=4.5 kg which in turn pushes against the block B of mass mB=7.5 kg, as shown in the figure. If there is no friction between the blocks and the horizontal surface, what is the force exerted on block B by block A? (a) 16 N (b) 20 N (c) 25 N (d) 35 N (e) 18 N

Solution Steps

First, we need to find the total mass of the system, which is the sum of the masses of block A and block B.

\[ m_{\text{total}} = m_A + m_B = 4.5 \, \text{kg} + 7.5 \, \text{kg} = 12.0 \, \text{kg} \]

Next, we use Newton's second law to find the acceleration of the entire system. The applied force \( F_{\text{app}} \) is 25 N.

\[ F_{\text{app}} = m_{\text{total}} \cdot a \]

Solving for \( a \):

\[ a = \frac{F_{\text{app}}}{m_{\text{total}}} = \frac{25 \, \text{N}}{12.0 \, \text{kg}} = 2.0833 \, \text{m/s}^2 \]

Now, we need to find the force that block A exerts on block B. This force is responsible for accelerating block B.

\[ F_{A \rightarrow B} = m_B \cdot a \]

Substituting the values:

\[ F_{A \rightarrow B} = 7.5 \, \text{kg} \cdot 2.0833 \, \text{m/s}^2 = 15.625 \, \text{N} \]

Rounding to the nearest whole number, we get:

\[ F_{A \rightarrow B} \approx 16 \, \text{N} \]

Final Answer