Question

a 0.50 - kilogram frog is at rest on the bank surrounding a pond of water. as the frog leaps from the bank, the magnitude of the acceleration of the frog is 3.0 meters per second². calculate the magnitude of the net force exerted on the frog as it leaps. show all work, including the equation and substitution with units. 2

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.

Step2: Substitute the given values

Given \(m = 0.50\space kg\) and \(a=3.0\space m/s^{2}\). Substitute into the formula: \(F=(0.50\space kg)\times(3.0\space m/s^{2})\)

Answer:

\(1.5\space N\)