Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides

Question

Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor.

a) If Zak’s speed is 3.00 when he starts to slide, what distance will he slide before stopping? d=1.84

b) Now, suppose that Zak’s younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak’s). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 over a distance of 1.00 . If her mass is 20.0 , what distance does she slide after Zak’s push ends? Remember that the frictional force acts on Greta during Zak’s push and while she is sliding after the push.

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Ngọc Khuê 5 years 2021-08-01T05:17:04+00:00 1 Answers 648 views 0

Answers ( )

  1. thucuc
    1
    2021-08-01T05:18:36+00:00
    Reply

    a) 1.84 m

    b) 1.55 m

    Explanation:

    a)

    In this problem, the only force acting on Zak along the direction of motion (horizontal direction) is the force of friction, which is

    F_f=-\mu mg

    where

    \mu=0.250 is the coefficient of friction

    m is Zak’s mass

    g=9.8 m/s^2 is the acceleration due to gravity

    According to Newton’s second law of motion, the net force acting on Zak is equal to the product between its mass (m) and its acceleration (a), so we have

    F=ma

    Here the only force acting is the force of friction, so this is also the net force:

    -\mu mg = ma

    Therefore we can find Zak’s acceleration:

    a=-\mu g=-(0.250)(9.8)=-2.45 m/s^2

    Since Zak’s motion is a uniformly accelerated motion, we can now use the following suvat equation:

    v^2-u^2=2as

    where

    v = 0 is the final velocity (he comes to a stop)

    u = 3.00 m/s is the initial velocity

    a=-2.45 m/s^2 is the acceleration

    s is the distance covered before stopping

    Solving for s,

    s=\frac{v^2-u^2}{2a}=\frac{0^2-3.0^2}{2(-2.45)}=1.84 m

    b)

    In this second part, Zak gives a push to Greta.

    We can find Greta’s velocity after the push by using the work-energy theorem, which states that the work done on her is equal to her change in kinetic energy:

    (F-F_f)d =\frac{1}{2}mv^2-\frac{1}{2}mu^2

    where

    F = 125 N is the force applied by Zak

    d = 1.00 m is the distance

    F_f=\mu mg is the force of friction, where

    \mu=0.250

    m = 20.0 kg is Greta’s mass

    g=9.8 m/s^2

    v  is Greta’s velocity after the push

    u = 0 is Greta’s initial velocity

    Solving for v, we find:

    v=\sqrt{\frac{2(F-\mu mg)d}{m}}=\sqrt{\frac{2(125-(0.250)(20.0)(9.8))(1.00)}{20.0}}=2.76 m/s

    After that, Zak stops pushing, so Greta will slide and the only force acting on her will be the force of friction; so the acceleration will be:

    a=-\mu g = -(0.250)(9.8)=-2.45 m/s^2

    And so using again the suvat equation, we can find the distance she slides after Zak’s push ends:

    s=\frac{v'^2-v^2}{2a}

    where

    v = 2.76 m/s is her initial velocity

    v’ = 0 when she stops

    Solving  for s,

    s=\frac{0-(2.76)^2}{2(-2.45)}=1.55 m

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