Answer with Explanation:
We are given that mass of an elevator filled with passengers =[tex]1.70\times 10^3 Kg[/tex]
a.a=[tex]1.2 m/s^2[/tex]
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
[tex]F_net=\sum F=ma[/tex]
[tex]F-mg=ma[/tex]
Substitute the values then we get
[tex]F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N[/tex]
Hence, the tension in the cable supporting the elevator=[tex]1.87\times 10^4 N[/tex]
b.t=8.5 s
We have to find the tension in the table during this time.
Fnet=0
F=W=[tex]1.7\times 10^3\times 9.8=1.67\times 10^4 N[/tex]
Hence, the tension in the cable during this time =[tex]1.67\times 10^4 N[/tex]
c. Deceleration=[tex]0.6 m/s^2[/tex]
t=3 s
We have to find the tension in the cable during deceleration.
[tex]F-mg=-ma[/tex]
[tex]F=mg-ma=m(g-a)[/tex]
[tex]F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N[/tex]
Hence, the tension in the cable during deceleration=[tex]1.56\times 10^4 N[/tex].
d.We have to find that the height of elevator moved above its original staring point .
We have to find the final velocity.
[tex]y_1=ut_1+\frac{1}{2}a_1t_1^2[/tex]
[tex]y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m[/tex]
Final velocity
[tex]v_1=a_1t_1[/tex]
[tex]v_1=(1.2)(1.5)=1.8 m/s[/tex]
[tex]y_2=v_1t_1=(1.8)(8.5)=15.3 m[/tex]
The third distance [tex]y_3[/tex] is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by
[tex]y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m[/tex]
The total distance moved by elevator from its original position
[tex]d=1.35+1.8+2.7=19.35 m[/tex]
Final velocity=0
