A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attach

Question

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator. ) What does the balance read when the elevator is ascending with a constant speed of 33 m/s? N (b) What does the balance read when the elevator is descending with a constant speed of 37 m/s? N What does the balance read when the elevator is ascending at 28 m/s and gaining speed at a rate of 10 m/s2? N Suppose that from t = 0 to t = 4.6 s, the elevator ascends at a constant speed of 10 m/s. Its speed is then steadily reduced to zero during the next 3.9 s, so that it is at rest at t = 8.5 s. Describe the reading of the scale during the interval 0

Explanation / Answer

a) constant speed means acceleration is zero.

a = 0

using Fnet = ma

T - 2g = 0

T = 2 x 9.81 = 19.62 N

b) still constant speed means a = 0

T = 19.62 N

c) a = 10 m/s^2 upwards

T - 2g = 2a

T= 2 x (9.81 + 10) = 39.62 N

d) 0 to 4.6 sec ,

constant speed means a = 0

T = 19.62 N

4.6 to 8.5 sec ,
now acc is donwards

a = 10 / 3.9 = 2.56 m/s^2

mg - T = ma

T =2 (9.81 - 2.56)

T =14.49 N

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