Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coefficient of f

Question

Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coefficient of friction between the sled and the snow is µ = 0.025.

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Huyền Thanh 4 years 2021-07-27T01:58:47+00:00 1 Answers 10 views 0

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  1. thachthao
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    2021-07-27T02:00:23+00:00
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    Answer:

    Explanation:

    The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton’s Second Law that says that sum of the forces acting on an object is equal to the object’s mass times its acceleration. For us, that looks like this because of the friction working against the sled:

    F – f = ma but of course it’s much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

    f = μF_n where μ is the coefficient of friction, and F_n is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

    F_n=mg so

    F_n=(105)(9.8) so the weight of the sled is

    F_n= 1.0 × 10³ with the correct number of sig dig there. Now to find f:

    f = (.025)(1.0 × 10³) so

    f = 25 to the correct number of sig fig. Now on to our “real” equation:

    F – f = ma and

    230 – 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

    230 – 25 will round to the tens place giving us 210. Then

    210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

    a = 2.0 m/sec²

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