an elevator filled with passengers has a mass of 1.70\times 10^3~\text{kg}1.70×10 3 kg. after starting to move up the elevat
or accelerates at a rate of 0.600~\text{m/s}^20.600 m/s 2 for 3.00 s. what is the tension in the cable during deceleration?
1 answer:
Answer:
15.64 KN
Explanation:
Given
Mass of the elevator = 1.70 × 10³ kg = 1700 kg
acceleration of the elevator = - 0.6 m/s²
Tension in the cable = ?
The free body diagram is given in the attached image
The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.
Force balance,
ma = T - mg
1700 (-0.6) = T - (1700×9.8)
T = 16660 - 1020 = 15640 N = 15.64 KN
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