Question

1. when the frictionless system shown above is accelerated by an applied force of magnitude the tension in the string between the blocks is (a) f (b) 2/3 f (c) ½ f (d) 1/3 f

Explanation:

Step1: Calculate the acceleration of the system

According to Newton's second law \(F = ma\), for the whole system \(m = m_1 + m_2=1 + 2=3kg\), and \(F\) is the applied force. So \(a=\frac{F}{m}=\frac{F}{3}\).

Step2: Analyze the force on the \(1 - kg\) block

Let the tension in the string be \(T\). For the \(1 - kg\) block, using \(F = ma\) (where \(m = 1kg\) and \(a=\frac{F}{3}\)), we have \(T=m_1a\). Substitute \(m_1 = 1kg\) and \(a=\frac{F}{3}\) into the formula, we get \(T=\frac{1}{3}F\).

Answer:

D. \(1/3F\)