A particle of mass m=5.00 kg is released from point A with an initial speed of 5

Question

A particle of mass m=5.00 kg is released from point A with an initial speed of 5m/s and slides on a frictionless track. ( A roller coaster )

A roller coaster with mass 5 m for point A, 3.20 m for point B, and mass of 2.00 m for point C

A. Determine the particle 's speed at point B ( use conservation of energy )

B. Determine the particle's speed at point C. same as above

C. Determine the net work done by the gravitational force in moving the particle from A to C.

Please show your work clearly and write step by step solution including numeric substitutions and etc. To make it easier, please do this in paper and include the pictures. thanks

Explanation / Answer

KE1 + PE1 = KE2 + PE2 ( energy conservation)

A) 1/2*5*5^2 + 5*9.8*5 = 1/2*5*V^2 + 5*9.8*3.2

=> V^2 = 60.28 m/s

=> V = 7.76 m/s

B).

1/2*5*5^2 + 5*9.8*5 = 1/2*5*V^2 + 5*9.8*2

=> V^2 = 83.8

=> V = 9.15 m/s

C) work done by the gravitational force = (-mg)*(-H) = -5*9.8*(2-5) = 147 Joule ( force is acting in -Y, displacement is -H)

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