more physics help needed?

Question

The instruments attached to a weather balloon have a mass of 4.6 kg. The balloon is released and exerts an upward force of 100 N on the instruments.
(a) What is the acceleration of the balloon and instruments?
Magnitude
____ m/s2

(b) After the balloon has accelerated for 10 s, the instruments are released. What is the velocity of the instruments at the moment of their release?
Magnitude
_____ m/s

(c) What net force acts on the instruments after their release?
Magnitude
______ N

(d) When does the direction of their velocity first become downward?
_____ s (after release)

Answer 1

(a) The force exerted by the balloon is counteracted by gravity and as these act directly opposite each other, then

F = force on instruments and balloon = (100 - mg) N

where g = 9.8 m/s^2

Use Newton's second Law

F = mass * acceleration = ma

here m = 4.6 kg, F = (100-mg) N = (100 -mg) kg/m^2

(note it is always a good idea that you are using the same units in a physics equation)

a = F/m = 100 / 4.6 - g = 21.74 - 9.8 m/s^2 = 11.94 m/s^2

b) Assuming "released" means the instructions are cut loose from the balloon to fall back to earth

v = acceleration * time + starting velocity = at +v0

As it started from rest, v0 = 0 and so

v = 11.94 * 10 = 119.4 m/s

c) Only gravity acts on the instruments so F = mg

= 4.6 * 9.8 N = 45.08 N

d) The same equation can be used as in b) but remember the acceration is negative and acts to slow the instruments down
, and the direction of velocity will change after it becomes zero

So using v = at + v0

0 = -gt + 119.4 and so t = 119.4/g = 119.4/9.8 = 12.08s (after release)

Answer 2

a) ma = 110N - mg; a = 12 m/s2
b) v = 0 m/s at moment of release
c) weight, mg = 45 N
d) This is a rather odd question. The direction of the velocity becomes downward instantly upon release due to the force of gravity acting upon objects.

Answer 3

these are good questions - look like fun to figure out - got no time now though, but you can do it - not too hard.