Question

1. the acceleration of the cart shown below is represented in the given graph. if a second block is added to the cart, what might be the resulting acceleration? 0 m/s/s 0.27 m/s/s 0.4 m/s/s 0.53 m/s/s

Explanation:

Step1: Recall Newton's second law

$F = ma$, where $F$ is force, $m$ is mass and $a$ is acceleration. Assume the force acting on the cart remains constant. Let the initial mass of the cart be $m_1$ and acceleration be $a_1$, and the new - mass after adding the block be $m_2=m_1 + m_{block}$.

Step2: Express the force before and after

Before adding the block, $F = m_1a_1$. After adding the block, $F=m_2a_2=(m_1 + m_{block})a_2$. Since $F$ is constant, $m_1a_1=(m_1 + m_{block})a_2$. Then $a_2=\frac{m_1a_1}{m_1 + m_{block}}$. We know that $a_2\lt a_1$ because the denominator $m_1 + m_{block}\gt m_1$. If we assume the block has a non - zero mass and the initial acceleration $a_1$ is non - zero, the new acceleration will be smaller. Without knowing the exact masses, we can still reason that the acceleration will decrease. If we assume the block has a mass comparable to the cart's mass, say the mass of the block is equal to the mass of the cart ($m_{block}=m_1$), then $a_2=\frac{m_1a_1}{m_1 + m_1}=\frac{a_1}{2}$. If the initial acceleration is around $0.53\ m/s^2$ (assuming some non - zero initial acceleration from the context of the problem), when we double the mass, the new acceleration will be smaller. Among the options, the value that represents a decrease in acceleration is $0.27\ m/s/s$.

Answer:

$0.27\ m/s/s$