Question

1. two blocks are pushed along a horizontal frictionless surface by a force of 20 newtons to the right, as shown above. the force that the 2 - kilogram block exerts on the 3 - kilogram block is (a) 8 newtons to the left (b) 8 newtons to the right (c) 10 newtons to the left (d) 12 newtons to the right (e) 20 newtons to the left 11. when the frictionless system shown above is accelerated by an applied force of magnitude f, the tension in the string between the blocks is (a) 2f (b) f (c) (2/3)f (d) (1/2)f (e) (1/3)f

Explanation:

Question 10

Step1: Find the acceleration of the system

Using \(F = ma\), where \(F = 20N\), \(m=m_1 + m_2=3 + 2=5kg\).
The acceleration \(a=\frac{F}{m}=\frac{20}{5}=4m/s^{2}\)

Step2: Analyze the force on the \(2 - kg\) block

Let the force exerted by the \(3 - kg\) block on the \(2 - kg\) block be \(F_{32}\). For the \(2 - kg\) block, \(F_{32}=m_2a\) (using \(F = ma\)). Substituting \(m_2 = 2kg\) and \(a = 4m/s^{2}\), we get \(F_{32}=2\times4 = 8N\) to the right. By Newton's third law, the force exerted by the \(2 - kg\) block on the \(3 - kg\) block \(F_{23}\) is equal in magnitude and opposite in direction to \(F_{32}\). So \(F_{23}=8N\) to the left.

Step1: Find the acceleration of the system

The total mass \(m=m_1 + m_2=1+2 = 3kg\). Using \(F=ma\), the acceleration \(a=\frac{F}{m}=\frac{F}{3}\)

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

Let the tension in the string be \(T\). For the \(1 - kg\) block, \(T=m_1a\). Substituting \(m_1 = 1kg\) and \(a=\frac{F}{3}\), we get \(T = 1\times\frac{F}{3}=\frac{F}{3}\)

Answer:

A. 8 newtons to the left

Question 11