A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

YumYum247

  • one year ago

Help me Please!!!

  • This Question is Closed
  1. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  2. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Determine the velocity of the elevator 12 sec after the passenger have entered the elevator

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @paki

  4. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    idk if i know use the formula....Vf = Vi X at because idk if the elevator was initailly at resting position.... -_-

  5. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @MAEMAEHOCKEY

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so i think it would be 500 N

  8. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did you get applied force 1.90 X 10^3N?!?!?! UP

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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]

  10. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok ya

  12. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  13. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @4n1m0s1ty

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i tried sorry i am not that much help

  16. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Its ok dear. i appreciate every effort you made into helping me :)

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.