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2 years ago

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The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from

rest. assuming no friction, calculate the speed at points 2, 3, and 4.

1 answer:

kirill [66]2 years ago

7 0

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3 </span>= 26)

Let’s use the formula:

v2= sqrt[2*Gravity*h1] where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]

= √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )

=√2*9.81m/s2*(45-26)

=√2*9.81m/s2*19

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]

=√2*9.81m/s2*(45-16)

=√2*9.81m/s2*(29)

v4=23.85m/s

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