A 9.47 kg particle is subjected to only conservative forces and its potential en

Question

A 9.47 kg particle is subjected to only conservative forces and its potential energy diagram is shown below Use the diagram to find the maximum velocity of the particle between points G and K. At point D the particle has a speed of 13.7 m/s 1100 900 800 700 Potential Energy (J)800 500 300 200 100 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 x position (m) Between points G and K, at what location is the particle moving the fastest? H O The particle is moving at the same speed at all of the points listed here SCROLL DOWN FOR MORE

Explanation / Answer

at D, PE = 200 J  

KE = 9.47 x 13.7^2 /2 = 888.7 J  

total energy = PE + KE = 1088.7 J  

so velocity will be maximum when PE is minium.

that is at point J.

Ans: J

at this point, PE = 100 J

KE = 1088.7 - 100 = 988.7 J  

988.7 = 9.47 v^2 / 2

v = 14.45 m/s

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