F = net force acting on the elevator in upward direction = 3000 N
m = mass of the elevator = 1200 kg
a = acceleration of the elevator = ?
Acceleration of the elevator is given as
a = F/m
a = 3000/1200
a = 2.5 m/s²
v₀ = initial velocity of the elevator = 0 m/s
Y = displacement of the elevator = 15 m
t = time taken
Using the kinematics equation
Y = v₀ t + (0.5) a t²
15 = (0) t + (0.5) (2.5) t²
t = 3.5 sec
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Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
Answer:
1/2 M V^2 = .1 M g H where 10% of PE goes into KE
V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2
V = 31.1 m/s increase in speed during descent
1 km / hr = 1000 m / 3600 sec = .278 m/s
V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr
