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52.(52G) A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed of the crate?

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  1. find the final speed at the end of the first 15 m...

    vf^2=2ad where a=force/mass
    solve for Vf. Then use that as Vi here

    vf^2=vi^2+2ad where a= (F-mu*110g)/mass
    and d is again 15m
    solve for vf.

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    bobpursley
  2. 7. ( 11E ) a) Find the momentum of an automobile of mass 2630 kg traveling 21.0 ms .b) Find the velocity ( in km/h ) of a light auto of mass 1170 kg so that it has the same momentum as the auto in part ( a ).

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  3. From a to b
    Wg= Fx cos 90 = 0
    Wn= Fx cos 90 = 0
    Wnet=Efk - Eik. Which is = to 1/2mVf^2 -1/2 mVi^2.
    Fxcos(0)= Efk. Cos of 0 =1. Vi =0 so use Efk
    Fx=1/2mVf^2
    2Fx/m = Vf^2
    Vf^2= square root of2Fx/m
    Vf= 9.77m/s
    X = distance, m= mass, F= force, V= velocity
    Efk = final kinetic energy and is 1/2 mv^2
    Eik= initial kinetic energy
    Fn= normal force, Fg, force gravity
    B to C
    Fg=0
    Fn= 0
    Wnet = Efk - Eik
    Fxcos(0) + Ffrxcos(180). Cos of zero =1 cos of 180 =-1
    Fr= 0.30(9.8)(110)(15)(-1). =-4851
    Fx - 4851 = Efk - Eik
    350 (15)-4851 = Efk - Eik
    399= 1/2mVc^2 - 1/2 m Vb^2
    399 = 1/2(110)Vc^2 -1/2(110)(9.8)^2
    399 = 1/2(110)Vc^2 -5282
    399-5282 = 55 Vc^2
    399 + 5282/55 = Vc^2
    Vc = square root of 399 + 5282/55
    Vc=10.2m/s

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