Question

a box with a mass of m is placed on a frictionless table, as shown below. block on frictionless table

Explanation:

Step1: Calculate net force for Trial 1

Assume \(F_1\) is the left - ward force and \(F_2\) is the right - ward force. Net force \(F_{net}=F_2 - F_1\) (taking right as positive). For Trial 1, \(F_1 = 20N\), \(F_2=10N\). But wait, no, according to Newton's second law \(F = ma\). Given \(m = 2kg\), \(a = 5m/s^2\). Using \(F=ma\), \(F=(2)(5)=10N\). Also, \(F_{net}=|F_1 - F_2|\), if \(F_1>F_2\), \(F_{net}=20 - 10=10N\)

Step2: Calculate net force for Trial 2

Using \(F = ma\), with \(m = 2kg\) and \(a = 15m/s^2\). Then \(F=(2)(15)=30N\). Also, \(F_{net}=|F_1 - F_2|\), since \(F_1 = 40N\), \(F_2 = 10N\), \(F_{net}=40 - 10=30N\)

Answer:

For Trial 1, the net force is \(10N\). For Trial 2, the net force is \(30N\)