A5.5 kg ball initially at rest is dropped from the top of a 12 m building. it hits the ground 1.75s later. find the net external force on the falling ball

The distance an object falls from rest in some length of time is

D = (1/2) (acceleration) (time)²

We know the height of the building and the time spent falling, so we can use this formula to find the acceleration of the ball:

12 m = (1/2) (acceleration) (1.75 sec)²

Multiply each side by 2:

24 m = (acceleration) (1.75 sec)²

Divide each side by (1.75 sec)² :

(24 m) / (1.75 sec)² = acceleration.

Now that we know the acceleration of the ball during the fall, we can use Newton's 2nd law of motion to calculate the net force on the ball:

F = m A

Force = (mass) (acceleration)

Force = (5.5 kg) (24 m) / (1.75 sec)²

Force = (5.5 x 24) / (1.75)²  Newtons

Force = 43.1 Newtons

The thing that makes this problem so interesting is that the ball does NOT fall with the acceleration of gravity.  Its acceleration is only 7.84 m/s², so it must be falling in some mysterious circumstances that really resist its downward motion.  Maybe it's falling through very very very thick fog, or through molasses, or thick chicken soup.  Or maybe it's hanging from a little tiny parachute, or there's a bunch of bees under it that really really do NOT want it to fall and are trying to hold it up.  Maybe the ball is actually falling on Venus, where the gravitational acceleration is 8.87 m/s² and there's some more resistance due to the extra-soupy atmosphere on Venus. Whatever it is, its rate of acceleration is substantially less than Earth's 9.8 m/s².

current flows in a wire

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j