sure, as it drops it picks up speed. The loss in height potential energy, m g h, turns into kinetic energy, (1/2) m v^2
In other words:
(1/2) m v^2 = m g h
v^2 = 2 g h
v = sqrt(2gh)
which is something a lot of us remember.
Note that the mass does not matter. In a vacuum, a feather falls as fast as a rock. (In air of course the drag of the air slows the feather down, but that comes in your aerodynamics course next semester)
From your 1500 Joules I guess you are using 10 m/s^2 for g and not 9.8 or 9.81 so I will use
g = 10 m/s^2
v = sqrt (20*3)= 2 sqrt 15 = 7.75 m/s
Please help me solve this, most likely simple, problem. I am just starting to learn physics and am finding it very confusing. If you could include a key to explain any formulas used it would be very helpful thank you.
I have calculated GPE to be 1500J but I am totally stuck, I think I need to transfer GPE to KE, but I don't really know how or even how it would help.
When the mass(50kg) is 3m off the ground the rope breaks and the
mass falls. At what speed does it hit the ground?
3 answers
Of course there is a more basic way to look at this.
F = m a = -mg
so
a = - g = -10
so v = -gt = -10 t
so+- initial speed * t - (1/2)(10) t^2
0 = 3 + 0 - 5 t^2
t = sqrt(3/5)
so
v = -10 sqrt(3/5) = -sqrt(60)
= -2 sqrt 15 = -7.75 which we knew
F = m a = -mg
so
a = - g = -10
so v = -gt = -10 t
so+- initial speed * t - (1/2)(10) t^2
0 = 3 + 0 - 5 t^2
t = sqrt(3/5)
so
v = -10 sqrt(3/5) = -sqrt(60)
= -2 sqrt 15 = -7.75 which we knew
Thank you very much, with a little googling I can follow along with your first answer. Thank you very much for your help.
1 answer