Find the answer for number 25.

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

momentum of soft ball three times less than baskey ball