QUESTION 8 2 points Save A A 2000 kilogram roller coaster car moves from point A
ID: 1784792 • Letter: Q
Question
QUESTION 8 2 points Save A A 2000 kilogram roller coaster car moves from point A, to point B, to point C along frictionless track shown in figure below. At point A the car had the speed of 20 m/s. Round your answers to the nearest integer. 15 m 25 m What is the change (final minus initial!) in the car's gravitational potential energy as the car moves from A to B? What is the change in the car's gravitational potential energy as the car moves from B to C? What is the change in the car's gravitational potential energy as the car moves from A to C? What is the change in the cars kinetic energy as the car moves from A to B? What is the change in the car's kinetic energy as the car moves from B to C? What is the change in the car's kinetic energy as the car moves from A to C? What is the change in the car's mechanical energy as the car moves from A to B? What is the change in the car's mechanical energy as the car moves from B to C? What is the change in the car's mechanical energy as the car moves from A to C?Explanation / Answer
Given
mass of car m = 2000 kg ,
initial speedof the car at A is 20 m/s
the poits are at a level from the ground is A---25m , B ---40 m , c--0 m
the tota energy of car at A is E = p.e+k.e = mgh + 0.5*mv^2 = 2000*9.8*25 + 0.5*2000*20^2J = 890000 J
the change in gravitational potential energy when it reaches from A - B is
DP.e == P.e_B - p.e_A = mg(h2-h1) = 2000*9.8(40-25) J = 294000 J
smae way the change in k.e is E = Dp.e + Dk.e ==> Dk.e = E - Dp.e = 890000 -294000 J = 596000 J
From B -- C
the change in gravitational potential energy when it reaches from B - C is
DP.e == P.e_C - p.e_B = mg(h2-h1) = 2000*9.8(0-40) J = -784000 J
the change in k.e is E = Dp.e + Dk.e ==> Dk.e = E - Dp.e = 890000 -(784000) J = 106000 J
the change in gravitational potential energy when it reaches from A - C is
DP.e = P.e_C - p.e_A = mg(h2-h1) = 2000*9.8(0-25) J = -490000 J
chang ein k.e from A - c is Dk.e = E- dp.e = 890000 - 490000 = 400000 J
change in mecha ical energy
mechanical energy is sum of k.e and p.e
from A - B
change in m.e = E -E_A-B = E - change in P.e + change in k.e = 890000 - (294000 + 596000) = 0 J
From B - C
change in m.e = E - (change in k.e + change in p.e)_B-C = 890000-( 106000 + 784000) = 0 J
From A - C
change in m.e = E - (change in k.e + change in p.e)_A-C = 890000-( 400000 +490000) = 0 J
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