...is accelerated upwrd at a rate of 3m/s^2?
a.1.39*10^5
b.2.31*10^5
c.2.42*10^5
d.2.61*10^5
2006-12-14 12:05:04 · 3 answers · asked by Chelsea 1 in Science & Mathematics ➔ Physics
The tension will equal the downward force which is
F=m*a
where a=9.81+3
solve for mass as
2*10^5/9.81
F=2/9.81*(9.81+3)*10^5
d is your answer
j
2006-12-14 12:19:23 · answer #1 · answered by odu83 7 · 1⤊ 0⤋
If M is the mass of the elevator,
The tension to move it up with an acceleration a is
T = M ( g + a) = M 12.8
M = 2*10^5 /9.8
T =2*10^5 ( 12.8 /9.8) = 2.61*10^5 N
2006-12-14 12:29:12 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
The tension on the cable when the elevator at rest is 2*10^5 N.
The mass of the elevator is
m = 2*10^5 N / g = 2*10^5 N / 9.8 m/s^2 = 0.204*10^5
The additional tension of the cable
F = ma = (0.204*10^5)(3 m/s^2) = 0.61*10^5
So the tension on the cable is d. 2.61*10^5
2006-12-14 12:23:32 · answer #3 · answered by Matisse 2 · 2⤊ 0⤋