An elevator filled with passengers has a mass of 1.70×103kg. (a) the elevator accelerates upward from rest at a rate of 1.20m/s2 for 1.50 s. calculate the tension in the cable supporting the elevator. (b) the elevator continues upward at constant velocity for 8.50 s. what is the tension in the cable during this time? (c) the elevator decelerates at a rate of 0.600m/s2 for 3.00 s. what is the tension in the cable during deceleration? (d) how high has the elevator moved above its original starting point, and what is its final velocity?

Answer with Explanation:

We are given that  mass of an elevator filled with passengers  =

a.a=

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second  law,the net force is given by

Substitute the values then we get

Hence, the tension in the cable supporting the elevator=

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W=

Hence, the tension in the cable during this  time =

c. Deceleration=

t=3 s

We have to find the tension in the cable during deceleration.

Hence, the tension in the cable during deceleration=.

d.We have to find that the height  of elevator moved  above its original  staring point .

We have to find the final velocity.

Final velocity

The third distance  is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

The total distance moved by elevator from its original position

Final velocity=0

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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