If the cart starts from rest at 52 meters above ground, what is the speed of the coaster when it gets to 13 meters. (assume that all PE is converted to KE)..... Any ideas on how to do this and show work for it.
2006-12-18 10:31:49 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
potential energy = mgh
with m = mass
g = 9.8 m/s^2
h = height
kinetic energy = 0.5 mv^2
with v = velocity
chance in PE = 9.8 x 39 x m
=> 0.5 m v^2 = 9.8 x 39 x m
=> v^2 = 2 x 9.8 x 39
=> v = sqrt(764.4) = 27.65 m/s2
2006-12-18 10:47:49 · answer #1 · answered by mitch_online_nl 3 · 0⤊ 0⤋
The difference in potential energies at the two points is
Mg (H1 - H2).
This energy is converted in to kinetic energy of the body = 0.5 M V*V
0.5 V*V = g(H1 - H2). (M cancels)
V*V = 2 g (H1 - H2) = 2*9.8*(52-13) = 764.4
V= 27.65m/s
2006-12-18 15:40:49 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
No need to mention energy in this problem, except to state that none is converted to heat.......
V = √(2g∆h) = √(2*9.8*(52-13)) = 27.65 m/s
QED
2006-12-18 15:31:21 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
Your skill power is mgh. you could desire to be responsive to how briskly you're going earlier the drop to calculate your kinetic power and the centripetal rigidity up there. The substitute in power (neglecting friction losses, which you will no longer do, yet you will besides except they provide you friction coefficients) is 0.
2016-12-18 15:39:44 · answer #4 · answered by starich 4 · 0⤊ 0⤋
v=sqrt(2gh)
Kinetic energy is 1/2*m*v^2
Potential energy is m*g*h
for complete conversion:
KE=PE
v=sqrt(2*g*h)
for your specific question:
g=9.81 m/s^2
h=(52-13)m
v=27.7m/s
j
2006-12-18 10:42:52 · answer #5 · answered by odu83 7 · 0⤊ 0⤋