Question

Katie (m = 45kg) rides in a roller coaster on a frictionless track. She is sitting on a bathroom-type scale (a 'normalometer'). The track has a loop-the-loop (radius r = 10m) as shown. The cart has a speed of 15m/s at the top of the inside of the loop and a speed of 20m/s at the bottom of the inside of the loop. a. What is Katie's weight? b. What is Katie's apparent weight (normal force) at the top of the loop? c. What is Katie's apparent weight (normal force) at the bottom of the loop?

Answer 1

Katie's weight is 441 N. Her apparent weight at the top of the loop is 1143 N, and at the bottom of the loop is 1713 N.

Step-by-step explanation:

To calculate Katie's weight, we can use the equation w = mg, where w is the weight, m is the mass, and g is the acceleration due to gravity (g = 9.8 m/s2). Katie's weight is therefore equal to (45 kg)(9.8 m/s2) = 441 N.

To find Katie's apparent weight at the top of the loop, we can consider the forces acting on her. At the top of the loop, the centripetal force provided by the normal force must be greater than the force of gravity. Therefore, the apparent weight is the sum of the gravitational force and the centripetal force, which can be calculated using the equation Fc = mv2/r, where Fc is the centripetal force, m is the mass, v is the velocity, and r is the radius. Plugging in the values, the apparent weight at the top of the loop is (45 kg)(15 m/s)2/(10 m) + (45 kg)(9.8 m/s2) = 1143 N.

Similarly, to find Katie's apparent weight at the bottom of the loop, we can use the same equation for the centripetal force. Plugging in the values, the apparent weight at the bottom of the loop is (45 kg)(20 m/s)2/(10 m) + (45 kg)(9.8 m/s2) = 1713 N.