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A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°

Calculate the
(a) gravitational potential energy at point A.

(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.

c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s

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A 900 kg roller coaster car starts from rest at point A rolls down the track goesaround a loop points B and C and then flies off the inclined part of the track class=

Respuesta :

Huntcrothersm

Answer:

gravitational potential energy at point A.

A) The gravitational potential energy at point A is; 705600 J

B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s

A) Formula for gravitational potential energy is;

PE = mgh

At point A;

mass; m = 900 kg

height; h = 80 m

Thus;

PE = 900 × 9.8 × 80

PE = 705600 J

B) Kinetic energy of the roller coaster at point C is given as;

KE = PE - W

We are given Workdone; W = 264870 J

Thus;

KE = 705600 - 264870

KE = 440730 J

Thus, velocity at point C is gotten from the formula of kinetic energy;

KE = ½mv²

v = √(2KE/m)

v = √(2 × 440730/900)

v = 31.295 m/s

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