An elevator filled with passengers has a mass of 1643 kg.(a)The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 3.00 s. Calculate the tension in the cable (in N) supporting the elevator. N(b)The elevator continues upward at constant velocity for 8.92 s. What is the tension in the cable (in N) during this time? N(c)The elevator decelerates at a rate of 0.600 m/s2 for 6.00 s. What is the tension in the cable (in N) during deceleration? N(d)How high has the elevator moved above its original starting point, and what is its final velocity? (Enter the height in m and the final velocity in m/s.)height mfinal velocity m/s

(a) To calculate the tension in the cable during the acceleration, we first need to calculate the force of gravity acting on the elevator. This is given by the equation F = m*g, where m is the mass of the elevator and g is the acceleration due to gravity (9.8 m/s^2).

F = 1643 kg * 9.8 m/s^2 = 16101.4 N

Next, we calculate the force required to accelerate the elevator upward. This is given by the equation F = m*a, where a is the acceleration of the elevator.

F = 1643 kg * 1.20 m/s^2 = 1971.6 N

The total force (or tension) in the cable is the sum of these two forces.

Tension = 16101.4 N + 1971.6 N = 18073 N

(b) When the elevator is moving at a constant velocity, the only force acting on it is the force of gravity. Therefore, the tension in the cable is equal to the weight of the elevator, which is 16101.4 N.

(c) When the elevator is decelerating, the force required to slow it down is given by F = m*a.

F = 1643 kg * 0.600 m/s^2 = 985.8 N

However, because the elevator is moving upward, the force of gravity is acting in the opposite direction. Therefore, the tension in the cable is the difference between the weight of the elevator and the decelerating force.

Tension = 16101.4 N - 985.8 N = 15115.6 N

(d) To calculate the total distance the elevator has moved, we need to calculate the distance it moved during each phase of its motion.

During the acceleration phase, the distance is given by d = 0.5at^2, where t is the time.

d = 0.5 * 1.20 m/s^2 * (3.00 s)^2 = 5.4 m

During the constant velocity phase, the distance is given by d = vt, where v is the velocity. The velocity at the end of the acceleration phase is v = at = 1.20 m/s^2 * 3.00 s = 3.6 m/s.

d = 3.6 m/s * 8.92 s = 32.112 m

During the deceleration phase, the distance is again given by d = 0.5at^2.

d = 0.5 * 0.600 m/s^2 * (6.00 s)^2 = 10.8 m

The total distance is the sum of these three distances.

Total distance = 5.4 m + 32.112 m + 10.8 m = 48.312 m

The final velocity after deceleration is given by v = u - a*t, where u is the initial velocity at the start of the deceleration phase. This is the same as the velocity at the end of the acceleration phase, which is 3.6 m/s.

Final velocity = 3.6 m/s - 0.600 m/s^2 * 6.00 s = 0 m/s

So, the elevator has moved 48.312 m above its original starting point and its final velocity is 0 m/s.