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Kinetic Energy and Work/ Work depends on acceleration?

  1. Oct 5, 2016 #1
    1. Problem Statement:

    A worker lifts a heavy crate vertically 1.0 m in 2.0 s. If she lifts the same crate the same distance in 1.0 s, the work was done by her is


    • A

      the same.



    • B

      two times as great.



    • C

      half as great.



    • D

      four times as great.



    • E

      one-fourth as great.



    • F

      zero.

    2. Relevant equations
    delta x=a.t^2/2+v0.t
    Fnet=ma Weight=mg
    W=F.d
    g=9.8N/kg
    3. The attempt at a solution
    v0=0m/sec
    delta x=a.t^2/2
    t=2sec
    1=a.4/2
    a=0.5m/sec^2
    F=m(0.5+9.8)=10.3m N
    W=10.3m J

    t=1sec
    1=a1/2
    a=2 m/sec^2
    F=m(2+9.8)=11.8m N
    W=11.8m J
    then work change. How the correct answer stays the SAME I didn't get it.
     
    Last edited by a moderator: Apr 14, 2017
  2. jcsd
  3. Oct 5, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Unfortunately, the problem statement is not clear regarding the acceleration during the lifting. You have assumed constant acceleration during the entire distance. Your work looks correct for this assumption .

    However, the person who constructed the problem was presumably assuming that the object is lifted at constant speed during essentially the entire distance. There would need to be acceleration at the very beginning to get the object moving at the required speed, but this acceleration is assumed to be over such a small time interval that the distance moved during the acceleration is negligible.
     
  4. Oct 5, 2016 #3
    it's obvious that acceleration differs since we are cutting the same distance but in lesser time. I'm really confused about it. If you have any other way tell me since I' not sure of mine.
     
  5. Oct 5, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Imagine lifting a box from the floor to a table top. Suppose you want the time it takes to be 20 seconds (a rather long time, but chosen to make a point). The natural way to do this would be for you to lift it at constant speed for the entire "trip" except at the very beginning and the very end. Very few people would lift it so that it has a constant acceleration during the entire displacement. So, the work done would essentially be the work done by the applied force to lift the box at constant speed. The initial acceleration phase only lasts for a very small time interval, and the work done during this small time interval can be neglected.

    If you reduce the time from 20 seconds to 10 seconds, the initial acceleration phase would have a different acceleration or a different time of acceleration (or both). Nevertheless, the work done during the acceleration interval is still negligible compared to the work done while the box is being lifted at constant speed.
     
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