The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon is exhibited on a platform 5.00m above the ground. After the exhibition, the watermelon is allowed to slide to the ground alone a smooth ramp. How high above the ground is the watermelon at the moment its kinetic energy is 4.61 kJ. What is its speed at this point?
I known the height is 1.01 m, however I need help finding the speed at this point!
Thank you! :)
I known the height is 1.01 m, however I need help finding the speed at this point!
Thank you! :)
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use KE = 1/2 m v^2 solve for v
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as initial velocity is zere v is just the velocity at the point where the total KE is 4610 J, and here we're only interested in the magnitude of the velocity or speed
4610= 1/2(118) v^2
v^2 = 4610(2)/118
v^2 = 78.14
v = sqrt (78.14) or 8.84 m/s
4610= 1/2(118) v^2
v^2 = 4610(2)/118
v^2 = 78.14
v = sqrt (78.14) or 8.84 m/s
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