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Work/Conservation of Mechanical Energy

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    An 80-N box is pulled 20m up a 30 degree incline by an applied force of 115 N that points upward, parallel to the incline. If the coefficient of kinetic friction between box and incline is 0.22, calculate the change in the kinetic energy of the box.
    2. Relevant equations
    ΔKE = 1/2mVf2-1/2mVi2 = mghi- mghf
    3. The attempt at a solution
    This is what i have understood so far, if there's a mistake please help me.

    Given:
    Fg= 80N so m=8.15kg
    hi= 20m
    θ= 30°
    μk= 0.22
    Fapp= 115N

    I drew the ramp, and labeled all the forces. Also, I found Fgx to be 40N. My question is, what do they mean by change in KE? Should I find work done? If so, is it Wnet= Fdcosθ ? I'm really confused, I've spent more than an hour looking for solutions online, but I don't understand the explanation and I decided to try asking myself. Please explain what I should do, I've tried solving several times, but I can't seem to find the answer, and I also have a lot of other problems to solve.

    Help is appreciated. Thank you
     
  2. jcsd
  3. Mar 7, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    What is the initial kinetic energy?
    What is the final kinetic energy?
    The difference is the change in kinetic energy.

    Can you list all types of energy involved? Where does the box gain energy, where does it lose energy?
    You did not introduce d. The formula can be relevant but I would choose a different approach.
     
  4. Mar 7, 2016 #3
    actually you should draw a free body diagram and find the net force which will be responsible for motion along the ramp-in up direction.then you can calculate the acceleration - one can assume initial kinetic energy to be zero and the potential energy change will be due to displacement along the ramp -
     
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