Respuesta :
Answer:
The answers to the question are
a. Before the elevator starts to move the spring scale registers 755.37 N
b. During the first 0.900 s the spring scale registers 847.77 N
c. While the elevator is traveling at constant speed the spring scale registers the force due to gravity on the man = 755.37 N
d. During the time it is slowing down the spring scale registers 706.45 N
Explanation:
We solve each part as follows, knowing the velocity and time during each phase of the elevator motion
a. Before the elevator starts to move the spring scale registers the weight of the man which is
Weight = mass × gravity, where gravity or g = 9.81 m/s² and the mass of the man = 77.0 kg
Hence the scale registers 77 × 9.81 = 755.37 N
(b) During the first 0.900 s the spring scale registers
The wight of the man ad the force of the upward moving elevator on him thus we are required to calculate the acceleration of the elevator thus
v = u + at where v = final velocity = 1.08 m/s u = initial velocity = 0 and t = 0,900 s therefore,
1.08 = 0.9×a or a = 1.08/0.9 = 1.2 m/s²
Hence the force on the man by the elevator plus the weight of the man = 755.37 N + 1.2×77 N = 847.77 N
(c) While the elevator is traveling at constant speed the spring scale registers the force due to gravity on the man or the weight of the man thus
77 × 9.81 = 755.37 N
(d) During the time it is slowing down the spring scale registers
Since in this case v = u-at then
a = 1.08/1.7 = 0.635
Hence the spring registers 77 ×(9.81 - 0.635) = 706.45 N
