what is the differnce between a force acting upon an object and the net force acting on an object?

Answer 1

The first is just one force acting upon the object (ex, your hand pushing a ball across the carpet); the second is the total of all forces acting upon an object (ex, your hand pushing the ball, the floor pushing up, gravity pulling down, drag from the carpet and air, etc).

Answer 2

A force acting on an object is referring to one and only one force that is acting on the object. For instance, if I try to push my car with 250 N of force, I am acting on the object with 250 N of force. Net force is the sum of all individual forces acting on an object. To back to my car, it does not move when I push on it. This is because friction is acting with 250 N of force in the opposite direction that I am pushing the car, giving me a net force of zero (no acceleration).

Answer 3

I always feel that net force concept must be clearly imbibed therefore, i drop this example everywhere
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Suppose m is the mass of body moving under the influence of a set of forces (with magnitudes and directions) as specified.
e.g. let us visualize a RAINDROP

(1) F1= mg, acting downwards
(2) F2= up thrust due to medium (buoyancy - cloud)
.............and volume of drop = vol*(rho-m)*g
(3) F3=drag of medium (stroke's force) upwards
.............6 pi *eta*r* velocity

Now you have to decide where the expected movement of m is desired (e.g. raindrop cannot go up). Thus, take cloud as ORIGIN and expected direction of fall as (+x).

Of course, m will be tested for its accelerated motion downwards so READILY prepare your

[m*d^2x/dt^2] downwards (positive, ready to lead m with acceleration x-double DOT)
>> it just like the probe-end of a radiologist performing ultrasound of pregnancy.

All F1, F2 and F3 >> now become subjugate to it [ ], and will be treated with + or minus relative to [ ] positive. In other words,

(+ F1), (- F2) and (- F3) all downwards

so NET FORCE = [m*d^x/dt^2] = + F1 - F2 - F3

The drop will fall when m*d^x/dt^2 = net positive or F1>(F2+F3)
F2 up thrust depends on variation in geometrical properties (vol) in nature - a limited variant. It so happens that stroke's drag (ETA-viscosity coeff remains stubborn in vapor phase - varies odd powers of temp than phase changed water) has to be cut off so that F3 falls drastically. Raindrops show us the positive net force downwards.

For clear picture of motion, writing a similar differential equation is a must.

When net force on m is =0 then m*d^x/dt^2=0
dx/dt =constant

if m*d^x/dt^2=0 after time "t" (or during transit) then net force (F1, F2, F3) has algebraically forced m to stop accelerating, thus acquire Terminal velocity.

you have to visualize this DE in every case.