PhanthomJay said:Apparently the radius of cuvature is 2 meters at the bottom of the loop, which you did not state. If so, your answer is correct, but they asled for the answer in terms of mg. Convert 31.25 m/sec^2 to how many 'g's , where g = ?).
The normal force at the bottom of a track is the force exerted by a surface on an object in contact with it, perpendicular to the surface. In the context of a track, it is the force that pushes the object in an upward direction to counteract the force of gravity pulling it down.
The normal force at the bottom of a track can be calculated using the formula: N = mg + ma, where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object on the track. This formula takes into account both the force of gravity and any additional acceleration that the object experiences on the track.
The normal force at the bottom of a track is affected by the mass of the object, the acceleration of the object, and the angle of the track. As the mass or acceleration of the object increases, the normal force will also increase. Additionally, a steeper angle of the track will result in a larger normal force.
The normal force at the bottom of a track is important because it is responsible for keeping the object on the track and preventing it from falling or sliding off. It also plays a crucial role in determining the speed and acceleration of the object on the track.
No, the normal force at the bottom of a track can never be negative. It is always directed perpendicular to the surface and therefore cannot have a negative value. If the normal force were to become negative, it would mean that the object is being pushed into the surface, rather than being supported by it.