Question

1. two forces are applied to a 2.0 - kilogram block on a frictionless horizontal surface, as shown in the diagram below. what is the acceleration of the block?

f₁ = 2.0n
2.0 kg
f₂ = 8.0n

frictionless surface

Explanation:

Step1: Calculate the net force

The net force \(F_{net}\) is the difference between the two forces since they act in opposite directions. \(F_{net}=F_{2}-F_{1}\).
Given \(F_{1} = 2.0\space N\) and \(F_{2}=8.0\space N\), so \(F_{net}=8.0 - 2.0=6.0\space N\) (in the direction of \(F_{2}\)).

Step2: Use Newton's second law \(F = ma\) to find acceleration

Newton's second law is \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. We need to solve for \(a\), so \(a=\frac{F_{net}}{m}\).
Given \(m = 2.0\space kg\) and \(F_{net}=6.0\space N\), then \(a=\frac{6.0}{2.0}\space m/s^{2}\).

Answer:

\(3.0\space m/s^{2}\) (in the direction of \(F_{2}\))