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Physics Classical Mechanics

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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)=
Question No. 2 any ideas.

  • Physics Classical Mechanics - ,

    for a: 41.67

  • Physics Classical Mechanics - ,

    for 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] [Note: v_i = 0 and t_i = 0]
    mv_f = (β/3)(t_f)^3 - (f_k)t_f
    v_f ≅ 10.83, but that's not the answer, so any suggeston any one???

  • Physics Classical Mechanics - ,

    Are you sure about a?

  • Physics Classical Mechanics - ,

    ellie, Thanks. I thought this is the most difficult question. I had trouble with these type of problems throughout the course. I do not have any suggestions. at this point.

  • Physics Classical Mechanics - ,

    I did pretty much the same of what you're showing and I got a) 25, b) 2.5 and c) 0.64 I haven't check it yet because it is my last submission and I want to be sure.

  • Physics Classical Mechanics - ,

    guys a is 25 for sure

  • Physics Classical Mechanics - ,

    have you check your data, because for a got a green mark, so A is ok, but b and c were not. There might be some where m = 3 and beta=1.2

  • Physics Classical Mechanics - ,

    For question 5 ballistic missile m1= 5*10^24kg; r1=6000km; m2<<m1; alpha=30 degrees; r2=(5/2)r1=15000; G=6.674*10^-11. What is the initial speed of the projectile? My solution:
    Vo^2=g*range/sin(theta)
    Vo^2=6.674*10^-11*(r1+r2/sin30degrees)
    Vo^2=(6.674*10^-11)*(21000/0.5)
    Vo^2=0.00000280308
    Vo=sqrt(0.00000280308)
    Vo=0.00167424
    Is this done correctly. Thanks for all your help.

  • Physics Classical Mechanics - ,

    so, no more suggestions for b and c of initial question? and Mets, check the videos from week 13.

  • Physics Classical Mechanics - ,

    Guys, please, can you show how you got 25 for a, because I got 41.6 as elli did. For b) my answer is same as ellis too. For c) I used formula for instantaneous power P=FV . So you just need to find values of F and V at t=4 and plug it in there.

  • Physics Classical Mechanics - ,

    there is different m & beta... no wonder if daoine get 25 for a. who has the formula for b & c?

  • Physics Classical Mechanics - ,

    q1 a)
    F = dp/dt
    dp = Fdt
    ∫dp = ∫Fdt
    p_f - p_i = ∫βt²dt
    p_f - 0 = (β/3)(t_f)^3 - (β/3)(t_i)^3

    no idea for part b) and c) any help ???

  • Physics Classical Mechanics - ,

    rickross, your solutions for a=41.6 which is comes in contrariety, to a=25.is it surerly right??

  • Physics Classical Mechanics - ,

    that's is grt formula i got it correct using that formula!
    hai Greco help for q8?

  • Physics Classical Mechanics - ,

    rickross still working on it i ll have it soon..

  • Physics Classical Mechanics - ,

    k thanks did u got q1) b ? and c? and if u got it help m ???

  • Physics Classical Mechanics - ,

    Anyone can explain the rocket question and the doppler shift question? I got half of each question correct and don't know how to move on.

  • Physics Classical Mechanics - ,

    answers depend on the values we have

    so, for A
    F = dp/dt -> dp = Fdt
    ∫dp = ∫Fdt
    p_f - p_i = ∫βt²dt
    p_f - 0 = (β/3)(t_f)^3 - (β/3)(t_i)^3

    for b:
    F - fk = m.a
    F - fk = m.dv/dt
    (F - fk)dt = m.dv
    ∫(F - f_k)dt = ∫m.dv
    m.vf - m.vi = (β/3)(tf)^3 - (fk)tf - [(β/3)(ti)^3 - (fk)ti] [Note: v_i = 0 and t_i = 0]
    m.vf = (β/3)(tf)^3 - (fk)tf

    for C) well, I guess we find vf for t=4 as in b, and then
    P=W/t -> (1/2m.vf^2)/t

    and for m=2 and β=1,
    A) 41.67
    B) like said before, I got 10.83 but it wasn't the right answer HELP!!!

  • Physics Classical Mechanics - ,

    For part A this is what I did and I got 20 for me and validated 25 for some of you.

    F=ma
    a=F/m
    Because this is not constant acceleration V=at^2. Do not divide by 2! Then P=mv which result in 25 for some of you and 20 for me.

  • Physics Classical Mechanics - ,

    Look at this however I am confused about the t^3. I know that is comes from integration but not sure of why the integration if the F(t)Bt^2 is supposed to describe you non linear force in function of a.
    answers yah_o_o c_om/
    question/
    index?qid=20131207034631AA7c7w6

  • Physics Classical Mechanics - ,

    from ELLI A its right, but just found;

    b) p=m*v, so
    v(t)=(integral Ff(t) from (the point where it starts to move) to t) /m

    the point where it starts to move:
    beta*t^2-m*g*mu=0
    t=sqrt(m*g*mu/beta)

    v(5)=[beta/3*5^3-m*g*mu*5-beta/3*sqrt(m*g*mu/beta)^3+m*g*mu*sqrt(m*g*mu/beta)]/m
    ----------------------
    c) P=F(4)*v(4)
    take F(t)=beta*t^2 and v(4) like in part b)

    beta=1 and m=2, uo=0.2
    B) vf=5.33
    C) P=??

  • Physics Classical Mechanics - ,

    I got 6.99 for part B. F=.8 and m=3. Would you please verify. Part A for me is 33.33
    Thanks.

  • Physics Classical Mechanics - ,

    Power should be the difference in KE when start moving and t final divided by time. Since KE and Power have the same unit and watts is a unit in function of time that is why you need to take the time interval of motion.

  • Physics Classical Mechanics - ,

    I got those new equations from
    jiskha . com / display.cgi?id=1386491909

    But I'm not sure about those, I'm still waiting for someone to confirm.... any

  • Physics Classical Mechanics - ,

    @Fima
    Hello fima... it seems to be the right equation, you see... The force only starts to push the block when it "wins" the Friction Force (Ff). So, its only after 2 seconds has passed. Using your numbers: Ff = m*g*mu = 2*10*0.2 = 4N... F=bt^2 = 1 * (2)^2=4N... So right after 2 seconds, F wins Ff, and the object starts to move on. So, you must integrate from 2 to 5 (question B, and 0 to 4 at C)... You would have:

    Fnet = ma
    Fnet = mdv/dt
    Fnet*dt = mdv, integrating from 2 to 5
    Fnet*(t_1-t_0)=mv_1 - mv_0
    Fnet*t_1 - Fnet*t_0=mv_1 - mv_0
    Fnet = F - Ff and v_0 = 0
    (F - Ff)*t_1 - (F - Ff)*t_0 = mv_1
    F*t_1 - Ff*t_1 - F*t_0 + Ff*t_0 = mv_1
    Changing F =b*t^2, and Ff= m*g*mu
    v_1 = [(b/3)*t_1^3 - m*g*mu*t_1 - (b/3)*t_0^3 + m*g*mu*t_0]/m
    Putting values in:
    v_1 = [(1/3)*5^3 - 2*10*0.2*5 - (1/3)*2^3 + 2*10*0.2*2]/2
    v_1 = 27/2 = 13.5 m/s. As far as I can tell. Don't know exactly how you got 5.33, did I do something wrong?

  • Physics Classical Mechanics - ,

    @Fima,
    You've said:
    B) vf=5.33
    But this v_f is only for t=4, that used at question C, at question B is used t=5. Check it out.

  • Physics Classical Mechanics - ,

    Thanks, That is v(4) for the next questions, but the procedure is correct

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