Question

90 kg man

1. a is standing in an elevator. calculate the apparent weight of the for the following

cases:

a. the elevator moves upward at a constant velocity.

b. the elevator accelerates downward at a rate 4 m/s².

c. the elevator accelerates upward at a rate 4 m/s².

d. the cable snaps and the elevator is in free fall.

Explanation:

Part a: Elevator moves upward at constant velocity

Step1: Recall Newton's laws (constant velocity means \(a = 0\))

The forces on the man are normal force \(F_N\) (upward) and weight \(mg\) (downward). By Newton's second law, \(\sum F = ma\). Since \(a = 0\), \(\sum F = F_N - mg = 0\).

Step2: Calculate weight and normal force

Mass \(m = 90\space kg\), \(g = 9.8\space m/s^2\). Weight \(mg = 90\times9.8 = 882\space N\). So \(F_N = mg = 882\space N\) (apparent weight is \(F_N\)).

Part b: Elevator accelerates downward at \(a = 4\space m/s^2\)

Step1: Apply Newton's second law (downward acceleration, so \(a\) is downward)

Forces: \(F_N\) (upward), \(mg\) (downward). Taking downward as positive, \(\sum F = mg - F_N = ma\).

Step2: Solve for \(F_N\)

Rearrange: \(F_N = mg - ma = m(g - a)\). Substitute \(m = 90\), \(g = 9.8\), \(a = 4\): \(F_N = 90\times(9.8 - 4)=90\times5.8 = 522\space N\).

Part c: Elevator accelerates upward at \(a = 4\space m/s^2\)

Step1: Apply Newton's second law (upward acceleration, upward is positive)

Forces: \(F_N\) (upward), \(mg\) (downward). \(\sum F = F_N - mg = ma\).

Step2: Solve for \(F_N\)

Rearrange: \(F_N = mg + ma = m(g + a)\). Substitute values: \(F_N = 90\times(9.8 + 4)=90\times13.8 = 1242\space N\).

Part d: Elevator is in free fall (\(a = g = 9.8\space m/s^2\) downward)

Answer:

s:
a. \(\boldsymbol{882\space N}\)
b. \(\boldsymbol{522\space N}\)
c. \(\boldsymbol{1242\space N}\)
d. \(\boldsymbol{0\space N}\)