Question

a 60 kg person is riding in an elevator. at time ( t_1 ), the elevator is accelerating downward with a magnitude of ( 2 , \text{m/s}^2 ). a short time later, at time ( t_2 ), the elevator is accelerating upward with a magnitude of ( 2 , \text{m/s}^2 ). the ratio of the normal force exerted by the elevator on the person at time ( t_1 ) to that at time ( t_2 ) is most nearly
a ( 2:3 )
b ( 4:5 )
c ( 5:6 )
d ( 1:1 )

Explanation:

Step1: Analyze the situation at \(t_1\)

When the elevator accelerates downward, using Newton's second law \(mg - N_1=ma\). Here \(m = 60\space kg\), \(a = 2\space m/s^{2}\), \(g= 10\space m/s^{2}\).
So \(N_1=m(g - a)=60\times(10 - 2)=60\times8 = 480\space N\)

Step2: Analyze the situation at \(t_2\)

When the elevator accelerates upward, using Newton's second law \(N_2 - mg=ma\).
So \(N_2=m(g + a)=60\times(10 + 2)=60\times12 = 720\space N\)

Step3: Calculate the ratio

The ratio \(\frac{N_1}{N_2}=\frac{480}{720}=\frac{2}{3}\)

Answer:

A. \(2:3\)