Question

4.10 q: a 25 - newton horizontal force northward and a 35 - newton horizontal force southward act concurrently on a 15 - kilogram object on a frictionless surface. what is the magnitude of the objects acceleration? (a) 0.67 m/s² (b) 1.7 m/s² (c) 2.3 m/s² (d) 4.0 m/s²

Explanation:

Step1: Calculate the net force

Since the forces are in opposite directions (northward and southward), we subtract them. Let \(F_1 = 25N\) (northward) and \(F_2=35N\) (southward). The net force \(F_{net}=F_2 - F_1\) (taking southward as positive).
\(F_{net}=35 - 25=10N\)

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

We know \(F = F_{net}=10N\) and \(m = 15kg\). To find \(a\), we rearrange the formula to \(a=\frac{F}{m}\).
\(a=\frac{10}{15}=\frac{2}{3}\approx0.67m/s^{2}\)

Answer:

A. \(0.67m/s^{2}\)