Physics

I was wondering if my answers are correct.

1. A 50kg person is taking a ride on an elevator travelling up at a steady speed of 2.5m/s. Find the time fo the elevator trip if the elevator does 4900 J of work on the person.

My answer: First I used the formula f=mg to find the force and got -490N. Then, I used the firmula w=fd and found the distance which is 10m. Lastly, I used v=d/t to find the time and got -4.0 seconds. Time shouldnt be negative but I got a negative number from the force equation.

2. A car slows from 108km/h to 90km/h over a course of 4.0s. THe force of the brakes was 1000N. What is the work done by the brakes on the car?

My answer: 110 000N

3. A pair of sleds weighs 25 kg and experience frictional force of 25N. If a dog team applies a 175 N force pulling it 12m find:
a) work done by friction
b) work done by dog team
c) net work done (using net force). how does the net work compare to answers from a and b
d) what form of energy does the frictional force represent? what form of energy does the net force produce?

My answers:
a) 300 N•m
b) 2100 N•m
c) net force=200N and net work=2400N•m. I'm not sure how it would compare to the previous answers.
d) frictional force represents thermal energy and net force produces kinetic energy?

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  1. 1. A 50kg person is taking a ride on an elevator travelling up at a steady speed of 2.5m/s. Find the time fo the elevator trip if the elevator does 4900 J of work on the person.

    My answer: First I used the formula f=mg to find the force and got -490N.
    ************
    - 490 N is gravity force DOWN on person.
    + 490 N is force elevator exerts UP on person
    *************************
    Then, I used the firmula w=fd and found the distance which is 10m.
    ***********
    YES
    ***********
    Lastly, I used v=d/t to find the time and got -4.0 seconds. Time shouldnt be negative but I got a negative number from the force equation.
    +++++++++++++++++++++
    Well my force is POSITIVE :)
    +++++++++++++++++++++++ So I get +4 sec

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  2. 2. A car slows from 108km/h to 90km/h over a course of 4.0s. THe force of the brakes was 1000N. What is the work done by the brakes on the car?

    My answer: 110 000N

    +++++++++++++++++++++++++++++++++++++++
    How far?
    In constant acceleration, average velocity is (final +initial)/2
    v = 99 *1000/3600 = 27.5 m/s
    goes 4 s * 27.5 m/s = 110 meters
    1000N * 110 m = 110,000 Joules
    So YES

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  3. a) 300 N•m YES (or JOULES)
    b) 2100 N•m YES (or Joules
    c) net force=200N and net work=2400N•m.
    ***************
    I get + for pull and - for friction so
    net = 175 - 25 = 150N

    Net work = 150 * 12 = 1800 Joules
    and
    1800 = 2100 -300 indeed
    *********************

    I'm not sure how it would compare to the previous answers.
    ******************
    useful work = work done - heat generated
    *********************
    d) frictional force represents thermal energy and net force produces kinetic energy? yes

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  4. Ohhh I see what I did wrong! Thanks a lot again!

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  5. for question number 1
    we need to find time but we cant find time if we don't have the distance. So first we will find distance.So apply a=Fnet/m ,
    a=0(speed is constant).
    0*50=Fnet=0N
    then as we have the mass the other force applying here is the gravitational force
    so Fnet=Fapp-Fg(gravitational force is acting in the opposite direction)
    We will find Fg=mg,
    Fg=50*9.8=490N
    so Fnet=Fapp-Fg
    0N=Fapp-490
    4900N=Fapp
    W=Fapp*d(work is the product of force applied and distance covered)
    4900=490d
    4900/490=d
    10m=d
    No we have distance and velocity we can easily find time
    v=d/t
    t=d/v
    t=10/2.5
    t=4seconds

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