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 second^2. Calculate The magnitude of the net force exerted on the frog as it leaps.

Answer 1

F = ma
F = (0.5 kg)(3 m/s^2)
F = 1.5 N

Answer 2

The magnitude of net force exerted on the frog is of 1.5 N.

Given data:

The mass of frog is, m = 0.50 kg.

The magnitude of acceleration of frog is, $a = 3.0 \;\rm m/s^{2}$.

As per the concept of Newton's Second law, the force applied on any object is equal to the product of mass and acceleration of an object. Then the expression is given as,

$F=m \times a$

Solving as,

$F = 0.50 \times 3.0\\\\F=1.5 \;\rm N$

Thus, we can conclude that the magnitude of net force exerted on the frog is of 1.5 N.

Learn more about the Newton's second law here:

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