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Therefore, $\vec{F} = m\vec{a}$ and $\displaystyle F= \frac{m(v_1 + v_2)}{t}$ are not the same thing; you are right to say that the former is the net force while the latter is the average force (when there is a collision, as that is how you derived the expression). Now, for your final question, there is not really such a thing as "average net ...
If one of the force is unknown, but its direction is known then you must know the acceleration in that direction in order to solve the above for the force component. Sometimes the above is treated as a static problem with $$\sum{\vec{F}_i} - m \, \vec{a}_{cm} =0$$ by including the inertial force in an opposite sense as a force in the free body ...
Suppose you carry out the rest of your calculation. Then you've found the net force in the z-direction in the north half of the sphere. However, the north half cannot exert any net force on itself, so this entire net force must be the same as the net force from the southern hemisphere.
After liftoff, the rocket experiences a net force: the force of thrust minus the force of gravity gives a positive net force upwards. However, at the moment of ignition, there are three forces: the normal, the gravity, and the thrust (let's assume the engine is at 100% immediately for simplicity).
The net-force causes net-work. So can we also state net-force causes acceleration in the direction of the net force and so also causes a change in kinetic energy of the system? (Asking this because I read a derivation of Bernoulli's equation, which can be taken as an equation of conservation of mechanical energy, from Newton's second law.
Does having a constant velocity always make the acceleration equal zero? For example: A 5 kg ball is moving at constant velocity of 15 m/s.
Ok so I am stuck on one of the problems about three blocks being pushed by a force: Three blocks on a frictionless horizontal surface are in contact with each other, as shown in Fig. 4–51. A for...
Force is a vector quantity. The first law talks of a single object and a force, without going in the details. A net force means a vector addition of forces, two equal and opposite forces add up to zero net force . This is expressed more clearly here.
The magnitude of the net force (Fnet) is calculated as √(FEM2 + FSM2), leading to a net force of approximately 4.81 × 1020 N. To find the direction of this net force, we use trigonometry to find the angle of the resultant force with respect to one of the given forces.
The distance between q1 and the other two charges is 2.00 m.So to determine the net force exerted on q1 from the other two charges, we know the individual charges between q1 and q2, and q1 and q3. We also know the angle, therefore we must use cosine law to solve for net force.