Pls explain in DETAIL how did you come up with the solution. I'm going to present it in class and idk how to explain the Equations 4.The mass of the three people, the car, radius of the loop of the roller coaster and the velocity at two different points aregiven. For calculation purposes we denote it as,ml = mass of Dahyun = 48.9 kgm2 =mass of Jeongyeon = 49. 1 kgm3 = mass of Chaeyoung = 46 kgmass of car M= 120 kgradius of the loop =r = 12 mspeed of the car at the bottom (point A) = va = 25 m/spoint A is at height = ha = 0speed of the car at the top (point B) = vb = 8 m/spoint B is at height = hb= 2r = 24 ma)The first part needs to be solved. In this part, we are asked to find the work due to friction as the car is moving frompoint A to B. By the work energy theorem, the amount of work done on an object is equal to the change in its kineticenergy.The force of gravity does positive work on an object moving down an inclined plane. The force of friction doesnegative work on a sliding object, thereby reducing its kinetic energy.work done by frictional force Wf = total change in energy= change in K.E+ change in P.E= 0.5x net mass (Va2 - Vb2) + mg(ha- hb)= 0.5 x(ml+m2+m3+M) (Va2 - Vb2 ) + (ml+m2+m3+M) g (ha- hb)=0.5x(48.9+49.1+46+120) x(252-82) + (48.9+49.1+46+120) x 9.8 (0-24)= 11959.2 Jg used here is acceleration due to gravityb)We have to find gravitational potential energy.at point A being at the bottom height ha 0at for point B at the top height hb = 2r = 24 mpotential energy is mghthen at point A = potential energy = P.E(A) = mgha = mgx0 = 0at point B = potential energy = P.E(B ) = mghb = (ml+m2+m3+M) g hb= (48.9+49.1+46+120) x 9.8 x24= 62092.8Jchange in potential energy = P.E(A) -P.E(B) = 0-62092.8 = -62092,8 J 4.The mass of the three people, the car, radius of the loop of the roller coaster and the velocity at two different points aregiven. For calculation purposes we denote it as,ml = mass of Dahyun = 48.9 kgm2 =mass of Jeongyeon = 49. 1 kgm3 = mass of Chaeyoung = 46 kgmass of car M= 120 kgradius of the loop =r = 12 mspeed of the car at the bottom (point A) = va = 25 m/spoint A is at height = ha = 0speed of the car at the top (point B) = vb = 8 m/spoint B is at height = hb= 2r = 24 ma)The first part needs to be solved. In this part, we are asked to find the work due to friction as the car is moving frompoint A to B. By the work energy theorem, the amount of work done on an object is equal to the change in its kineticenergy.The force of gravity does positive work on an object moving down an inclined plane. The force of friction doesnegative work on a sliding object, thereby reducing its kinetic energy.work done by frictional force Wf = total change in energy= change in K.E+ change in P.E= 0.5x net mass (Va2 - Vb2) + mg(ha- hb)= 0.5 x(ml+m2+m3+M) (Va2 - Vb2 ) + (ml+m2+m3+M) g (ha- hb)=0.5x(48.9+49.1+46+120) x(252-82) + (48.9+49.1+46+120) x 9.8 (0-24)= 11959.2 Jg used here is acceleration due to gravityb)We have to find gravitational potential energy.at point A being at the bottom height ha 0at for point B at the top height hb = 2r = 24 mpotential energy is mghthen at point A = potential energy = P.E(A) = mgha = mgx0 = 0at point B = potential energy = P.E(B ) = mghb = (ml+m2+m3+M) g hb= (48.9+49.1+46+120) x 9.8 x24= 62092.8Jchange in potential energy = P.E(A) -P.E(B) = 0-62092.8 = -62092,8 J

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