A man of 80kg stands on a spring scale in an elevator during the first 10seconds of motion from rest. the tension T in the cable is 9300N. find the reading R of the scale in newtons during the interval and upward velocity v on the elevator at the end of the 6seconds. the total mass of the elevator man and scale is 800kg.
Given:
Total mass (M) = 800 kg
Tension in the cable (T) = 9300 kg ms⁻²
Mass of Man (m) = 80 kg
* Then equilibrium of forces in vertical direction will be:
T-Mg=Ma
Where, g= acceleration due to gravity
and a= acceleration due to upward force(acceleration of elevator)
9300-(800×10)= 800a
a= 1.625 ms⁻²
* For Reading of the scale :
R= m× total acceleration
R= m(g+a)
R= 80(10+1.625)= 80×2.625
= 210 ms⁻² = 210Newtons
* For upward velocity after 6 seconds
final velocity= initial velocity +(acceleration×time)
v=u+at
v= 0+1.625×6 = 9.725 ms⁻¹
velocity at the end of six seconds will be 9.725 ms⁻¹
[Note: g=10ms⁻² was taken for simplicity. Different values will give different answers]
To find the reading R of the scale during the interval, we need to determine the net force acting on the man in the elevator.
1. First, we need to calculate the gravitational force acting on the man.
Gravitational force (Fg) = Mass (m) * Acceleration due to gravity (g)
Given that the mass of the man is 80 kg and the acceleration due to gravity is approximately 9.8 m/s²:
Fg = 80 kg * 9.8 m/s² = 784 N
2. Next, we need to consider the tension in the cable as the elevator moves. During the interval, both the gravitational force and the tension in the cable contribute to the net force.
Net force (Fnet) = Tension (T) - Gravitational force (Fg)
Given that the tension in the cable is 9300 N:
Fnet = 9300 N - 784 N = 8516 N
3. Since the reading R on the scale represents the net force acting on the man, the reading on the scale during the interval is 8516 N.
Now, let's calculate the upward velocity (v) of the elevator at the end of 6 seconds.
1. According to the problem, the elevator starts from rest, so the initial velocity (u) is 0 m/s.
2. The acceleration (a) of the elevator can be calculated using the equation of motion: v = u + at.
Rearranging the equation, we have a = (v - u) / t.
Substituting the values, we get a = (v - 0) / 6s = v/6.
3. The net force (Fnet) acting on the elevator can be calculated using Newton's second law: Fnet = mass (m) * acceleration (a).
Given that the mass of the elevator, man, and scale is 800 kg:
Fnet = 800 kg * (v/6) = 8516 N.
4. We can now solve for v:
800 kg * (v/6) = 8516 N.
v/6 = 8516 N / 800 kg.
v = (8516 N / 800 kg) * 6.
v ≈ 64 m/s.
Therefore, the reading R on the scale during the interval is 8516 N, and the upward velocity of the elevator at the end of 6 seconds is approximately 64 m/s.