What is the net force that acts on a 10-N falling object when it encounters 4-N of air resistance?

Answer 1

The FIRST thing one does to solve force problems (and weight is a force) is identify all the forces acting on the system in the problem (e.g., your falling object).

You have two forces: weight and drag (air resistance).

The SECOND thing one does is add up all the forces. This is how we find the "net" force (f). If we consider a downward net force as positive (f>0), then an upward net force must be negative (f<0). Similarly, if there are sideways forces (and there are none in your problem), we can say any to the left are positive and any to the right are negative; or vice versa. Which way is negative and positive makes no difference as long as we are consistent throughout the problem. The important thing is to ensure forces acting in opposite directions (left, right, up, down) have different signs (plus, minus).

Your net force (up and down only) = f = W - D; where W = mg = weight of the mass pulling downward on your falling object and D = drag force pulling upward on that same object. Thus, f = ma = 10 N - 4 N = 6 N. g ~ 10 m/sec^2 (actually averages around 9.81 m/sec^2 on Earth's surface).

Net force (f) determines if the falling object is still accelerating in its fall or if the velocity it has reached is constant (known as the terminal velocity). Thus, if f = ma > 0 (with down as a positive), then a > 0 and the falling object is still accelerating. But if f = ma = 0, the a = 0 and the object is no longer accelerating...it is at terminal velocity.

f = ma = 0 = W - D; so W = D when the object has reached terminal velocity. Drag force = D = 1/2 rho Cd A v^2; where rho is air density, Cd is a drag coefficient based on the shape and attitude of the object, A is the cross sectional area of the object, and v = the velocity of the falling object. When the object and air factors are known, we can use W = D to solve for v, which is the terminal velocity under these conditions.

Answer 2

Simple Simple.

10-4=6 N

Answer 3

6 N.

Have you learned to use free body diagrams? Those help a lot for questions like this. Ask your teacher or another student to explain them.

At the beginning of its fall (implying it is slow so we neglect air resistance for now) the only force acting on it is gravity. Its mass is 1 kg, and gravity pulls it with 10N of force, so it will begin to accelerate downward at 10 m/s^2 (force / mass = accleration).

At some speed its air resistance will have grown to 4N of air resistance. Now there are two forces acting on it: 10 N of gravity pulling it down and 4N of air resistance pushing up (air resistance is always in the direction opposite of velocity). The net force is now 6N downward. At this point the object will still be accelerating downward, but at a rate of 6 m/s^2.

Answer 4

a million. if 2 products of the same length fall via the air at diverse speeds, the only which encounters the in simple terms proper air resistance is the ______. [a] slower merchandise 2. on a lengthy alley a bowling ball slows down because it rolls. that is because _______ are pesent. [b] gravity and friction 3. at the same time as a junked motor vehicle is beaten right into a compact cube_______. [d] the mass, weight and quantity all replace. 4. A undergo weighs 3500 N grasps a vertical tree and slides down at consistent speed. The pressure of friction that acts on the undergo is ______. (Idk what this one is sorry) 5. In a vacum, a coin and a feather fall on the same fee because _____. [a] gravity isn't present 6. a tremendous and a small persn want to parachute at equivalent terminal velocities. the better individual may have to_______. [c] pull upward on the assisting strands to diminish the downward information superhighway pressure