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3 July, 13:37

Astrophysicist Neil deGrasse Tyson steps into an elevator on the 29th floor of a skyscraper. For some odd reason, there is a scale on the floor of the elevator. Neil, whose mass is about 115 kg, decides to step on the scale and presses the button for a lower floor. The elevator starts traveling downwards with a constant acceleration of 1.5 m/s2 for 6.0 seconds, and then travels at a constant velocity for 6.0 seconds. Finally, the elevator has an upward acceleration of 1.5 m/s2 for 6.0 seconds as it comes to a stop.

A. If each floor is approximately 4 m tall, which floor does the elevator stop at?

B. If the mass of the elevator is 1,200 kg, what is the maximum tension of the elevator cable?

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  1. 3 July, 17:02
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    A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)

    B. T = 14859.5 N = 14.89 KN

    Explanation:

    A.

    First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:

    s₁ = Vi t + (0.5) at²

    where,

    s₁ = distance covered during accelerated downward motion = ?

    Vi = initial speed = 0 m/s (since elevator is initially at rest)

    t = time taken = 6 s

    a = acceleration = 1.5 m/s²

    Therefore,

    s₁ = (0 m/s) (6 s) + (0.5) (1.5 m/s²) (6 s) ²

    s₁ = 4.5 m

    also we find the final velocity using 1st equation of motion:

    Vf = Vi + at

    Vf = 0 m/s + (1.5 m/s²) (6 s)

    Vf = 9 m/s

    Now, the second part of downward motion is with constant velocity. So:

    s₂ = vt

    where,

    s₂ = distance covered during constant speed downward motion = ?

    v = Vf = 9 m/s

    t = 6 s

    Therefore,

    s₂ = (9 m/s) (6 s)

    s₂ = 54 m

    Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:

    s₃ = s₁ = 4.5 m

    Therefore, the total distance covered by elevator is given by following equation:

    s = s₁ + s₂ - s₃ (Downward motion taken positive)

    s = 4.5 m + 54 m - 4.5 m

    s = 54 m

    Therefore, net motion of the elevator was 54 m downwards.

    So the final floor will be:

    Final Floor = Initial Floor - Distance Covered/Length of a floor

    Final Floor = 29 - 54 m/4m

    Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)

    B.

    The maximum tension will occur during the upward accelerated motion. It is given by the formula:

    T = m (g + a)

    where,

    T = Max. Tension in Cable = ?

    m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg

    g = 9.8 m/s²

    a = acceleration = 1.5 m/s²

    Therefore,

    T = (1315 kg) (9.8 m/s² + 1.5 m/s²)

    T = 14859.5 N = 14.89 KN
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