force acting on a mass of 50kg when it will will accel' it from 2m/s to 9m/s in 10s?.......I got 9-2 = 7, f=ma, (50x7) = 350n Is this right?
the one im stuck on though, is a ball of mass 5kg launched upwards with 800n force gets to height 'H'. What is 'H' and what is the time from launch until ball returns to ground. (when g = 9.8m's^2)
I know this doesn't quite belong in maths forum but you guys have ben a big help in the past, any input would be much appreciated. THANKS!
First a note: The unit "Newton" is named after someone, so it should properly be abbreviated with "N," not "n."
F = ma
800 N = (5 kg)*a
a = 800/5 m/s^2 = 160 m/s^2
The net acceleration is then
160 m/s^2 - 9.8 m/s^2 = 150.2 m/s^2 (upward)<-- Since the object's weight still acts on it.
So, what is the equation of motion of an object under a constant acceleration?
y = y0 + v0*t + (1/2)a*t^2
I'll set the origin at the point where the ball lifted off and +y upward. So y0 = 0 m and v0 = 0 m/s (presumably it was launched from rest, the problem never actually stated this.) So at the end of 20 s the height H will be:
H = (1/2)a*t^2
H = (1/2)*150.2*(20)^2 m = 30040 m.
To find out the time from launch to when it hits the ground we need two pieces of information: how long it takes for the ball to reach the ground from the height H, and what the velocity is at the end of the 800 N acceleration period. We'll do this part first.
While the ball is accelerated upward we have:
v = v0 + at <-- The ball starts from rest, so v0 = 0 m/s.
v = 150.2*20 m/s = 3004 m/s
(Note: since the sign on this velocity is + the velocity is upward.)
Now I'm going to reset the problem such that the time starts at the end of the 800 N acceleration phase. The ball starts at a height of 30040 m with an (upward) initial velocity of 3004 m/s and an acceleration only due to gravity of -9.8 m/s^2 (negative because it's acting downward.) The equation of motion is:
y = 30040 + 3004t - (1/2)*9.8*t^2
We wish to know when y = 0 m again. (Then add 20 s to this time and we get our final answer.) So:
-4.9t^2 + 3004t + 30040 = 0
Using the quadratic equation I get:
t = 622.903 s or t = -9.842 s
The negative time is clearly a ridiculous answer, so it takes 622.903 s to get back to the ground from this height.
Thus the total time of flight is
622.903 s + 20 s = 642.903 s
-Dan