Consider a 74.5-kg man standing on a spring scale in an elevator. Starting from

Question

Consider a 74.5-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.55 m/s in 0.575 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.85 s and comes to rest.

Part 1 of 5 -Conceptualize Based on sensations we experience riding in an elevator, we expect that the man feels slightly heavier when the elevator first starts to ascend, lighter when it comes to a stop, and his normal weight when the elevator is not accelerating. His apparent weight is registered by the spring scale beneath his feet, so the scale force measures the force he feels through his legs, as described by Newton's third law. Part 2 of 5-Categorize We draw a force diagram and apply Newton's second law for each part of the elevator trip to find the scale force. The acceleration is found from the change in speed divided by the elapsed time. ng Part 3 of 5- Analyze Consider the force diagram of the man shown as two arrows. The force Fs is the upward force exerted on the man by the scale, and his weight is Fg= mg kg)(9.80 m/s2) N. With +y defined to be upwards, Newton's second law gives the following equation, where Fs is the scale force, and a is the acceleration the man experiences as the elevator changes speed. So we have that the upward scale force is given by the following equation. Fs= mg + may (a) Before the elevator starts to move, ay - 0, so we have that Fs = mg = N.

Explanation / Answer

Fg = m*g
Fg = 74.5 * 9.8 N
Fg = 730.1 N

Fs = Fg + m*ay

Acceleration while ascending,
v = u + ay*t
1.55 = 0 + ay*0.575
ay = 2.70 m/s^2


Fs = Fg + m*ay
Fs = 730.1 + 74.5 * 2.70
Fs = 931.3 N

Acceleration while stopping,

v = u + ay*t
0 = 1.55 + ay*1.85
ay = - 0.837 m/s^2

Fs = Fg + m*ay
Fs = 730.1 - 74.5 * 0.837
Fs = 667.8 N

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