Answer :
Explanation:
It is given that, [tex]F_{w}[/tex] = 27.8 kN = [tex]27.8 \times 10^{3} N[/tex]
Now, formula for the mass of elevator is as follows.
[tex]F_{w} = mg[/tex]
m = [tex]\frac{F_{w}}{g}[/tex]
= [tex]\frac{27.8 \times 10^{3}}{9.8}[/tex]
= [tex]2.84 \times 10^{3}[/tex] kg
(a) When acceleration is increasing at a rate of 1.22 [tex]m/s^{2}[/tex]. Force according to the Newton's second law of motion is as follows.
F = T - mg = ma
[tex]T - 27.8 \times 10^{3} = 2.84 \times 10^{3} \times 1.22[/tex]
T = [tex]31.27 \times 10^{3} N[/tex]
Hence, tension in the cable if the cab’s speed is increasing at a rate of 1.22 [tex]m/s^{2}[/tex] is [tex]31.27 \times 10^{3} N[/tex].
(b) When acceleration is decreasing at a rate of 1.22 [tex]m/s^{2}[/tex]. Force according to the Newton's second law of motion is as follows.
F = T - mg = ma
[tex]T - 27.8 \times 10^{3} = 2.84 \times 10^{3} \times (-1.22)[/tex]
T = [tex]24.34 \times 10^{3} N[/tex]
Hence, tension in the cable if the cab’s speed is decreasing at a rate of 1.22 [tex]m/s^{2}[/tex] is [tex]24.34 \times 10^{3} N[/tex].