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$\bullet$ $\bullet$ A 350 kg roller coaster starts from rest at point $A$ and slidesdown the frictionless loop-the-loop shown in the accompany-ing figure. (a) How fast is this roller coaster moving at point $B ?$(b) How hard does it press against the track at point $B$ ?

(a) $v_{2}=15.96 \mathrm{m} / \mathrm{s}$(b) $n=11433.3 \mathrm{N}$

Physics 101 Mechanics

Chapter 7

Work and Energ

Physics Basics

Applying Newton's Laws

Kinetic Energy

Potential Energy

Energy Conservation

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problem number it d seven. In this problem, a roller coasters is given. This roller coaster also has a circular part. That one that initial ground level. It's like this. The radius off the circular part is six meters. The point A height is given us 25 meters. The point B height is well meters which is equal to the diameter of the circular part. This site is given us four meters. Oh, this is three meters. No, we have to find the velocity off the roller coaster car at point B to find this velocity off the car weevil, apply the energy conservation principle between the points A and me. Bye. Energy conservation. The initial energy off the car at the point here it was addressed is, um, J into height from the ground appointed, which is 25 meters. If frictional forces are neglected, then this gravitational potential energy is equal to the total energy off the car at B. That is a top most point off the circular part off the roller coaster. This is a little huh um, the square plus, um j and I return of the circular part which is well, meters. This will give the velocity guys. 15 born, 97 Meet that. What Second? No. For the second part of the question, we have to find the normal force on the surplus tech at the top. Most position that would be we have already seen that the radius off this circular part is it a little six meters at one Beat the weight off the car that's down works and the normal force from the track also ex downwards From the knowledge of circular motion, we know better fit object moves in a circular but then it is acted upon by force which ex towards the center of the circle. This force is called the centripetal force does. But in this case, the normal force and the gravity force will combine to provide the centripetal force through the car this weekend, right? And less mg is it called Judah Sent to people Force, which is equal to M v squared by, uh, Ari's are in this case is six meters, so normal force is equal to, um we square by six minus m g. The mass off the car is already given us 3 50 cages and we have calculated the velocity off the car at Point B in Party has 15.97 This is a squired upon sake minus 3 50 in tow, 9.8 just normal force will be 11 four for three. Fine. Here it Newton's. The concept off the centripetal force is very important to solder part B.

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