Q11: A roller-coaster car may be represented by a block of mass 50.0 kg. The car

Question

Q11: A roller-coaster car may be represented by a block of mass 50.0 kg. The car is released from rest at a height h = 57.0 m (point A) above the ground and slides along a frictionless track. The car encounters a loop of radius R 19.0 m at ground level. as shown. [Useg-9.8 m/s) (a) [4 pts] What is the total energy of the car at the point A, as it sits at rest? (b) [8 pts] Find the kinetic energy K of the car at the top of the loop (point B). (c) [4 pts] Find the kinetic energy K of the car at the bottom of the loop.

Explanation / Answer

here,

mass ,m = 50 kg

h = 57 m

R = 19 m

a)

the total energy of car at A , TEa = m * g * h

TEa = 50 * 9.81 * 57 J

TEa = 2.8 * 10^4 J

b)

using conservation of energy

the kinetic energy K of the car at the top , KE = m * g * ( h - 2R)

KE = 50 * 9.81 * ( 57 - 2 * 19) J

KE = 9.3 * 10^3 J

c)

using conservation of energy

kinetic energy gained = potential energy lost

the kinetic energy of the car at the bottom , KEf = m * g * h

KEf = 50 * 9.81 * 57 J

KEf = 2.8 * 10^4 J

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