Quote:A bike and its rider with a total mass of 100 kg initially moves horizontally with a velocity of 50m/s at an elevation of 600m above the ground where g=9.75 m/s^2. Determine a) the final velocity for a kinetic energy change of 500 kJ and b) the final elevation if the potential energy decreases by 500 kJ

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Please always start a new topic for posting your question,anyways here is what you have to do

For kinetic energy we have initial KE=1/2 mv^2= 0.5*100*50*50
Let final speed be S
Final KE

1/2*100*S^2 =50S^2

Change in KE = 50S^2-125000 =500000

Solve this for S

Similarly for PE we have initial PE=mgh =100*9.75*600=585000 j
let final height be L
Final PE = mgL
Change in PE = 585000-100*9.75*L=500000
Solve this for L to get answer

Hope this helps