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Work Problem in Watts

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data

    The maximum power your body can deliver in lifting an object vertically is 78 W. How fast could you lift, at constant speed, a 2L full water container?

    P=78W
    m=2kg
    t=?
    v=?


    2. Relevant equations

    P=ΔE/t
    E=1/2mv2

    3. The attempt at a solution

    78W=(.5*m*v2)/t

    t=(.5*2*v2)/78

    I am unsure what to do when I have a speed constant.
     
    Last edited: Oct 6, 2013
  2. jcsd
  3. Oct 6, 2013 #2
    Just remember that if you are lifting it at constant speed that means net force must be zero, or in other words your force should balance that of gravity. Then the rest is fairly straight forward calculus.

    W is work done.

    dW = F.dx
    P = dW/dt
    that implies P = F.(dx/dt)=F.v

    Oh, and welcome to pf!!
     
  4. Oct 6, 2013 #3
    I don't think I'm suppose to use calculus to solve this problem.

    However if W is equal to what is done by gravity and the formula for P=W/t, this is what I get:

    P=[itex]\frac{W}{t}[/itex] then t=[itex]\frac{m*g}{P}[/itex] which equals P=[itex]\frac{2*9.81}{78W}[/itex]=0.25s

    Is this what you're trying to imply?
     
  5. Oct 6, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Don't confuse Work (Joules) with Power (Watts). Sometimes the variable W is used for Work, but there is also a unit called a Watt, denoted by W, which is not the same thing. A Watt is a unit of power, and represents a Joule/second. The given value, 78 W, means that you can produce 78 Watts of power for the described maneuver.

    The current problem is asking you for "how fast" you can do something. In lay terms ("common language usage") that could mean in how short a time, but in physics "how fast" implies a speed. So you're looking for a speed. Look though your notes or text for a relationship between power and velocity... :wink:
     
  6. Oct 6, 2013 #5
    Ok so this is the relationship I found:

    P = [itex]\frac{E}{T}[/itex] = [itex]\frac{0.5*m*v^2}{t}[/itex]

    However, I do not have the t variable.

    So this is essentially what I can narrow the equation down to:

    78W = [itex]\frac{0.5*2*v^2}{t}[/itex] which equals to [itex]\frac{v^2}{t}[/itex]=78

    I am unsure what to do when I am that this step as the seconds in the time cannot cancel out with the speed.
     
  7. Oct 6, 2013 #6
    Never mind, I think I understand it now.

    With the equation P=[itex]\frac{E}{t}[/itex] the units are essentially [itex]\frac{kg*m^2}{s^3}[/itex]. This is equivalent to the product of F (N or [itex]\frac{kg*m}{s^2}[/itex]) and speed ([itex]\frac{m}{s}[/itex]). With this in my mind, the speed will be easy to determine as the force is known to be the product of mass and gravity.

    Is there any flaw in this explanation?
     
  8. Oct 6, 2013 #7

    gneill

    User Avatar

    Staff: Mentor

    Nope, that is well reasoned. In fact, a very useful relationship is the fact that the power delivered by a force F to a body moving at speed V is P = F*V :smile:
     
  9. Oct 7, 2013 #8
    yes, but the work done by gravity is m*g*h. so your expression becomes

    P=[itex]\frac{m*g*h}{t}[/itex]

    Here mg is the weight and h/t is the change in height with time, or simply your required velocity

    The only energy change is gravitational potential, since Kinetic Energy is constant due to velocity being constant.

    W = ΔE = m*g*h

    no, not at all but it could lead to more confusion in other situations. Like gneill said and I showed it above

     
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