KE =m•v²/2 =20•12.2²/2 =1488 J.
ΔPE = m•g•Δh = 20•9.8•20 = 3920 J.
Δ KE = ΔPE = 3920 J.
KE1 =KE + Δ KE = 1488 +3920= 5408 J.
The horizontal velocity isn’t changed. The vertical component is appeared, therefore, the velocity (as the vector sum of horizontal and vertical components) is increased.
A 20.0-kg model plane flies horizontally at a constant speed of 12.2 m/s.
(a) Calculate its kinetic energy.
. J
(b) The plane goes into a dive and levels off 20.0 m closer to Earth. How much potential energy does it lose during the dive? Assume no additional drag.
J
(c) How much kinetic energy does the plane gain during the dive?
J
(d) What is its new kinetic energy?
J
(e) What is its new horizontal velocity?
m/s
-i got at 1488.4
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