SOLVED: DIRECTION: Compute for Potential Energy PE = mgh m = 1 kg, g = 9.8 m/s^2, h = 44 m PE = 1 kg * 9.8 m/s^2 * 44 m m =
AK = -AU For gravitational field Let a ball of mass m is dropped from position (3) (as shown in figure) At point 1 PE = 0 KE = mu? v=J2gh so
SOLVED: known: P E=m g h =(705)(9.81)(50) =345.8 kJ K E=(1)/(2) m v^2=1 / 2(705)(1.00)^2=352.5 J P E=m g n=(705)(9.81)(0) P Ey B=0 J K E=1 / 2 m v^2 K E=P E=0 K E=1 / 2 m v^2 O=1 / 2 m v^2 V=0
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a) Find the PE at A PE = m g h = ( 500 kg )( 9.8 m/s2 )( 30 m ) - ppt download