A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is fri

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A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?

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  1. Answer:

    The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s

    Explanation:

    I’ll assume that the correct question is

    A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?

    mass of box = 51 kg

    for the first 12 m, it is pulled with a constant force of 240 N

    The acceleration of the box for this first 12 m will be

    from F = ma

    a = F/m

    where F is the pulling force

    m is the mass of the box

    a is the acceleration of the box

    a = 240/51 = 4.71 m/s^2

    Since the body started from rest, the initial velocity u = 0

    applying Newton’s equation of motion to find the final velocity at the end of the first 12 m, we have

    [tex]v^{2}= u^{2}+2as[/tex]

    where v is the final velocity

    u is the initial velocity which is zero

    a is the acceleration of 4.71 m/s^2

    s is the distance covered which is 12 m

    substituting value, we have

    [tex]v^{2}[/tex] = 0 + 2(4.71 x 12)

    [tex]v^{2}[/tex]  = 113.04

    [tex]v = \sqrt{113.04}[/tex] = 10.63 m/s

    For the final 10.5 m, coefficient of friction is 0.21

    from  f = μF

    where f is the frictional force,

    μ is the coefficient of friction = 0.21

    and F is the pulling force of the box 240 N

    f = 0.21 x 240 = 50.4 N

    Net force on the box = 240 – 50.4 = 189.6 N

    acceleration = F/m = 189.6/51 = 3.72 m/s^2

    Applying newton’s equation of motion

    [tex]v^{2}= u^{2}+2as[/tex]

    u is initial velocity, which in this case =  10.63 m/s

    a = 3.72 m/s^2

    s = 10.5 m

    v = ?

    substituting values, we have

    [tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)

    [tex]v^{2}[/tex]  = 112.9 + 78.12

    v  = [tex]\sqrt{191.02}[/tex]  = 13.82 m/s

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