Questions: Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP=9.3 N. Here A has a mass mA=10.6 kg and B has a mass mB=7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
(a) What is the magnitude (in m / s^2 ) of the acceleration of the two boxes?
□ m / s^2
(b) What is the force exerted on mB by mA? In other words what is the magnitude (in N ) of the contact force FC?
□ N
(c) If Alex were to push from the other side on the 7.0-kg box, what would the new magnitude (in N ) of FC be?
□
Transcript text: Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force $F_{P}=9.3 \mathrm{~N}$. Here $A$ has a mass $m_{A}=10.6 \mathrm{~kg}$ and $B$ has a mass $m_{B}=7.0 \mathrm{~kg}$. The contact force between the two boxes is $\vec{F}_{C}$. The coefficient of kinetic friction between the boxes and the floor is 0.04 . (Assume $\vec{F}_{\mathrm{P}}$ acts in the $+x$ direction.)
(a) What is the magnitude (in $\mathrm{m} / \mathrm{s}^{2}$ ) of the acceleration of the two boxes?
$\square$ $\mathrm{m} / \mathrm{s}^{2}$
(b) What is the force exerted on $m_{B}$ by $m_{A}$ ? In other words what is the magnitude (in $N$ ) of the contact force $\vec{F}_{C}$ ?
$\square$ N
(c) If Alex were to push from the other side on the $7.0-\mathrm{kg}$ box, what would the new magnitude (in $N$ ) of $\vec{F}_{C}$ be?
$\square$
Solution
Solution Steps
Step 1: Calculate the total mass of the system
The total mass \( m_{\text{total}} \) is the sum of the masses of both boxes:
\[ m_{\text{total}} = m_A + m_B = 10.6 \, \text{kg} + 7.0 \, \text{kg} = 17.6 \, \text{kg} \]
Step 2: Calculate the frictional force
The frictional force \( F_{\text{friction}} \) is given by:
\[ F_{\text{friction}} = \mu \cdot m_{\text{total}} \cdot g \]
where \( \mu = 0.04 \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ F_{\text{friction}} = 0.04 \cdot 17.6 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 6.9056 \, \text{N} \]
Step 3: Calculate the net force
The net force \( F_{\text{net}} \) is the applied force \( F_p \) minus the frictional force:
\[ F_{\text{net}} = F_p - F_{\text{friction}} = 9.3 \, \text{N} - 6.9056 \, \text{N} = 2.3944 \, \text{N} \]
Step 4: Calculate the acceleration of the system
The acceleration \( a \) is given by Newton's second law:
\[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{2.3944 \, \text{N}}{17.6 \, \text{kg}} = 0.136 \, \text{m/s}^2 \]
Final Answer
The magnitude of the acceleration of the two boxes is \( 0.136 \, \text{m/s}^2 \).