Mass of the body, m = 3 \(\text {kg}\)
Initial speed of the body, u = 2 \(\text m / \text s\)
Final speed of the body, v = 3.5 \(\text m / \text s\)
Time, t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
v = u + at
a = \(\frac{\text v-\text u}{\text t}\)
a = \(\frac{3.5-2}{25}\)
a = \(\frac{1.5}{25}\)
a = 0.06 \(\text m / \text s^2\)
As per Newton’s second law of motion, force is given as:
F = ma
F = 3 × 0.06 = 0.18 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ?
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:

Newton’s 3rd law states that there is an equal and opposite reaction for every action.