Question

5 stephen is riding an amusement park ride. his actual weight is 877 n at one point during the ride, stephen is accelerating downward at a rate of 7.75 m/s². what is stephens apparent weight during this period of acceleration? 184 n 1379 n 1570 n 226 n

Explanation:

Step1: Calculate Stephen's mass

We know that weight \(W = mg\), where \(W = 877\ N\) and \(g=9.8\ m/s^{2}\). So, \(m=\frac{W}{g}=\frac{877}{9.8}=89.5\ kg\)

Step2: Use Newton's second law \(F_{net}=ma\)

The net force \(F_{net}=mg - N\) (where \(N\) is the normal force, which is the apparent weight). Given \(a = 7.75\ m/s^{2}\) and \(F_{net}=ma\), we can rewrite the equation as \(N=m(g - a)\)
Substitute \(m = 89.5\ kg\), \(g = 9.8\ m/s^{2}\) and \(a=7.75\ m/s^{2}\)
\(N=89.5\times(9.8 - 7.75)=89.5\times2.05 = 184\ N\)

Answer:

184 N