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Momentum change

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A block of mass m=2 kg is initially at rest on a horizontal surface. At time t=0, we begin pushing on it with a horizontal force that varies with time as F(t)=Beta t^2, where Beta = 1.0 N/s^2. We stop pushing at time t1, 5s [F(t)=0 for t>1].
    (a) First, assume the surface is frictionless. What is the magnitude of the final momentum at t1=5s?
    P final(t=t1)=
    (b) lets now consider a new situation where the object is initially at rest on a rough surface. The coefficient of static friction is 0.2. What is the speed of the block at time t2=5s?
    v(t=t2)=
    (c) What is the power P provided by the force F(t) at time t3=4s (in Watts) i the case where there is friction (part(b))?
    P(t=t3)=
    2. Relevant equations
    Can you please check if my approach is right? I have some doubts.
    p=∫Fdt ,
    instantaneous power P=Fv
    3. The attempt at a solution

    part a) p=∫βt2=1/3βt3

    part b) F - f_k = ma
    F - f_k = mdv/dt
    (F - f_k)dt = mdv
    ∫(F - f_k)dt = ∫mdv
    mv_f - mv_i = (β/3)(t_f)^3 - (f_k)t_f - [(β/3)(t_i)^3 - (f_k)t_i] [v_i = 0 and t_i = 0]
    mv_f = (β/3)(t_f)^3 - (f_k)t_f

    part c)
    P=Fv
    to find v I used same approach as for b (but for t=4) and F=1*42
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your work looks good. You didn't include β in your expression for the answer in (a), but I guess that's because it has a numerical value of 1 N/s2.

    Of course you will need to plug in the time values and figure out the friction force to get your final numerical answers.

    Good work.

    [Welcome to PF!]
     
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