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A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 584 N. As the elevator later stops, the scale reading is 398 N. Assume the magnitude of the acceleration is the same during starting and stopping.

(a) Determine the weight of the person.
N

(b) Determine the person's mass.
kg

(c) Determine the magnitude of acceleration of the elevator.
m/s2

User CauseYNot
by
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1 Answer

3 votes

Answer:

(a) The weight of the person is 491 N

(b) The person's mass 50.1 kg

(c) The magnitude of acceleration of the elevator is 1.85 ms⁻²

Step-by-step explanation:

a)

when the lift starts moving then the apparent weight is R₁ = W + ma ........ (1)

when the lift ic going to rest, the apparent weight is R₂ = W - ma ......... (2)

add (1) and (2)

R₁ + R₂ = 2W

Given R₁ = 584N, and R₂ = 398N

584 + 398 = 2W

982 = 2W

491 = W

W = 491 N

b)

since W = mg

491 = m (9.8)

491 / 9.8 = m

m = 50.1kg

c)

From the equation (1), acceleration is

a = (R₁ - W) / m

a = (584 - 491) / 50.1

a = 1.85 ms⁻²

User John Spax
by
4.5k points