Final answer:
At the top of a vertical loop, a cart must travel at approximately 14.0 m/s for passengers to experience their normal weight. If this speed is maintained, passengers' apparent weight at the bottom of the loop would be 3g.
Step-by-step explanation:
The question asks for the required speed of a cart at the top of a vertical loop so that the passengers will feel their normal weight (scenario a), and then what their apparent weight would be at the bottom of the loop if that speed is maintained (scenario b).
For scenario a, at the top of the loop, passengers feeling their normal weight means the centripetal force required to keep them in circular motion is provided entirely by gravity. This means that the centripetal acceleration (ac) is equal to the acceleration due to gravity (g). The centripetal acceleration is given by the formula ac = v^2/r, where v is the speed and r is the radius of the loop. By setting ac = g, we can solve for v:
g = v^2/r, so v = √(gr).
Substituting g = 9.8 m/s^2 and r = 20.0 m, we find:
v = √(9.8 m/s^2 * 20.0 m)
v ≈ 14.0 m/s
For scenario b, at the bottom of the loop, the passengers will feel both the force of gravity and the centripetal force due to being in circular motion, which will result in a stronger net force and thus a greater apparent weight. The apparent weight can be larger than the real weight: apparent weight = mg + mv^2/r. Since v = 14.0 m/s and r = 20 m, we get an apparent weight equal to:
mg + m(14.0 m/s)^2 / 20.0 m = mg + 2mg = 3mg
This indicates that the passengers would feel an apparent weight equal to 3 times their normal weight, or 3g.