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. (a) What does the balance read when the elevator is ascending with a constant speed of 23 m/s? (b) What does the balance read when the elevator is descending with a constant speed of 28 m/s? (c) What does the balance read when the elevator is ascending at 27 m/s and gaining speed at a rate of 13 m/s2? (d) Suppose that from t = 0 to t = 4.3 s, the elevator ascends at a constant speed of 10 m/s. Its speed is then steadily reduced to zero during the next 3.7 s, so that it is at rest at t = 8.0 s. Describe the reading of the scale during the interval 0

Explanation / Answer

The free-body diagram shows the forces acting on the 2-kg block as the elevator ascends at a constant velocity. Because the acceleration of the elevator is zero, the block is in equilibrium under the influence of T and mg. Apply Newton’s second law of motion to the block to determine the scale reading.

(a) Apply Fy = may to the block to obtain

                 T – Fg = may

                  T – mg = may

For motion with constant velocity, ay = 0 and:    T – mg = 0 ==========> T = mg

                T = 2*9.8 = 19.6 N

(b) As in Part (a), for constant velocity, ay = 0. Hence:

                T – mg = 0 =====> T = mg = 19.6 N

(c)   T = m(g + ay) = 2(9.8 + 13) = 45.6 N

(d) During the interval 0 < t < 4.3 s, ay = 0. Hence:

                   T0-->4.3 = 19.6 N

Using its definition, calculate a for 4.3 s < t < 8.0 s:

                 a = v/t = (0 - 10)/3.7 = -2.7 m/s^2

                   T4.3-->8 = 2(9.8 – 2.7) = 14.2 N

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