We are given:
Mass of load = 140 kg
Acceleration = 0.4 m/s² downward (from Q24)
24. Finding the tension in the cable:
Ma = Mg - T
where 'a' is the acceleration of the load and T is the tension in the cable
replacing the variables, we get:
(140)(0.4) = (140)(9.8) - T
T = (140)(9.8) - (140)(0.4)
T = 1372 - 56
T = 1316N
25. Finding the distance covered when the load accelerates for 20 s:
in this case, the load will start from rest
so, initial velocity = 0 m/s
acceleration = 0.4 m/s²
from the second equation of motion:
s = ut + (1/2)at²
where s is the distance covered, u is the initial velocity and t is the time taken
s = (0)(20) + (1/2)(0.4)(20)²
s = 0 + 0.2*(400)
s = 80 m
here we didn't take the acceleration due to gravity of the mass into consideration because it is connected to the crane, which would act against any forces and since we are given that the total acceleration is 0.4 m/s², we will just assume that it's the final acceleration of the mass