Question

crate 1 of mass ( m_1 ) is connected to crate 2 of mass ( m_2 ) by a string with negligible mass. the crates are on a rough, horizontal surface. crate 2 is pulled to the right with constant acceleration by a force of constant magnitude ( f ). crate 1 experiences a friction force ( f_1 ) and crate 2 experiences a friction force ( f_2 ) both of which are less than ( f ). after a short period of time, the string breaks. the force ( f ) continues to be exerted on crate 2. which of the following are correct expressions for the magnitude ( a_1 ) of the acceleration of crate 1 and the magnitude ( a_{cm} ) of the acceleration of the two - crate systems center of mass after the string breaks?
a ( a_1=\frac{f_1}{m_1} ) and ( a_{cm}=\frac{f - f_2}{m_2} )
b ( a_1=\frac{f_1}{m_1} ) and ( a_{cm}=\frac{f - f_1 - f_2}{m_1 + m_2} )
c ( a_1=\frac{f - f_1}{m_1} ) and ( a_{cm}=\frac{f - f_2}{m_2} )
d ( a_1=\frac{f - f_1}{m_1} ) and ( a_{cm}=\frac{f - f_1 - f_2}{m_1 + m_2} )

Explanation:

Step1: Analyze the force on Crate 1

After the string breaks, the only horizontal force acting on Crate 1 is the friction force \(f_1\). According to Newton's second law \(F = ma\), for Crate 1, \(f_1=m_1a_1\), so \(a_1=\frac{f_1}{m_1}\).

Step2: Analyze the force on the two - crate system's center of mass

The total external force on the two - crate system is \(F - f_1 - f_2\) (because \(f_1\) acts on Crate 1 and \(f_2\) acts on Crate 2). The total mass of the system is \(m_1 + m_2\). According to Newton's second law for the center of mass \(F_{ext}=M_{total}a_{cm}\), where \(F_{ext}=F - f_1 - f_2\) and \(M_{total}=m_1 + m_2\), so \(a_{cm}=\frac{F - f_1 - f_2}{m_1 + m_2}\)

Answer:

B. \(a_{1}=\frac{f_{1}}{m_{1}}\) and \(a_{cm}=\frac{F - f_{1}-f_{2}}{m_{1}+m_{2}}\)