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An elevator that contains three people with the masses of 72Kg, 84Kg, and 35Kg respectively has a combined mass of 1030Kg. The cable attached to the elevator exerts an upward force of 1.20 X 10^4N but the frictional force opposing the motion of the elevator is 1.40 X 10^3N.
Determine the velocity of the elevator 12 sec after the passenger have entered the elevator
idk if i know use the formula....Vf = Vi X at because idk if the elevator was initailly at resting position.... -_-
F_a=1.90 x 10^3 N [up] F_f=1.40 x 10^3 N [down] F_net=1.90 e^3 N [up] + 1.40e^3 N [down] F_net= 500 N [up] a_net=F_net / m a_net=500 N [up] / 35 kg a_net= 0.49 m/s^2 [up]
so i think it would be 500 N
how did you get applied force 1.90 X 10^3N?!?!?! UP
F_a=1.20 x 10^4 N [up] F_g=mg F_g=(1030 kg)(9.8 m/s^2 [down]) F_g=1.01 x 10^4 N [down] F_net=F_a + F_g F_net=1.20 x 10^4 N [up] + 1.01 x 10^4 N [down] F_net=1.90 x 10^3 N [up]
Let's break down the question for a sec.... This is the first part of the question. and i answered it. 49b) Calculate the net acceleration of the elevator and its passangers. Ans) Given: Total mass = 1030Kg, Upward Force = 1.20 x 10^4N, Frictional Force = 1.40 X10^3N Required: Net Acceleration = ? Applied Force(Up) + Frictional Force(Down) = m. Net Acceleration 1.20 x 10^4N (Up) + (-1.40 X10^3N) (Down)) = 1030Kg. Net Acceleration 12000N(up) = 1030Kg. Net Acceleration (Divide both sides by 1030Kg) (10,600Kg.m/sec^2 (Up))/1030Kg = (1030Kg X Net Acceleration)/1030Kg 10.29m/sec^2 = Net Acceleration Therefore, the acceleration of the elevator with all three passengers on board is 10.29m.sec^2.
49d) Calculate the normal force acting on this passenger? Ans) In normal circumstances, the gravity is pulling against us with a force of 9.80kg.m/sec^2 or 9.80N but when the elevator itself goes at a net acceleration of 10.29m/sec^2, the person in the elevator with a mass of 35Kg will experience even greater gravitational force of attraction towards the elevator as he would on the surface of earth. Therefore, the normal force of the person would be significantly higher in the elevator as it’d be on the surface. Normal Force = m X Net Force of Acceleration Normal Force = 35Kg X 10.29m/sec^2 Normal Force = 360.15N or 360N
This is the last question. and i don't know how to solve it :( 49e) Determine the velocity of the elevator 12 sec after the passengers have entered the elevator.
i tried sorry i am not that much help
Its ok dear. i appreciate every effort you made into helping me :)