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Homework check

  1. Nov 17, 2004 #1
    1.A 25.0 kg object is attracted towards the earth by a force of gravity of 75.0 N. How far from the earths centre is it?

    my answer: g=Fg/m g=75N/25kg g=3 N/kg

    2. A 250 kg freezer is being pushed across the floor at a constant speed by a force of 750N. What is the coefficient of friction between freezer and floor?

    my answer: Fg=mg Fg=250kg(9.8N) Fg=2450kg/N
    m=Fapplied/Fg m=750N/2450kg/N m=0.31kg

    3. A piece cut from a bicycle inner tube is 0.70m long when it carries a load of 25.0 N the spring constant is 300 N/m. What will the length of the piece of rubber when the load is 75.0N?

    my answer: F=Kx F=300N/m / 75.0N F=4m in length

    4. A 60.0kg sled is coasting with a constant speed of 10.0m/s over smooth ice. It enters a 6m stretch of rough ice where the force of friction is 120N. With what speed does the sled emerge from the rough ice?

    my answer: m=Ffr/Fn m=120N/588N m=0.2kg

    5. A force of 5.0N gives a mass m1 and acceleration of 8.0m/s^2 and the same force gives mass m2 an acceleration of 24.0m/s^2. What acceleration would it give the two when they are fastened together?

    my answer: a=Fnet/m a=5.0N/0.833kg a=6.0m/s^2

    m1=(5.0N) / (8.0m/s^2) m1=0.625kg m2=(5.0N) / (24.0m/s^2) m2=0.208k

    plz tell me if i messed up anywhere
  2. jcsd
  3. Nov 17, 2004 #2


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    Hi physica.
    You don't seem to have answered the question here. You are asked for a distance and end with g=3 N/kg.
    g is the gravitational constant which is used on the surface of the earth.
    To find the approximate distance from the center of the earth, use Newton's law of gravity:

    Once again. You are not asked to give a mass, but a coefficient of friction.
    Use [itex]F_f=\mu N[/itex], where [itex]F_f[/itex] is the frictional force and N is the normal force.
    The unit of F is not meters...
    (One way to check whether you did it right is to ask yourself: "Does the answer I obtained make sense?")

    Use F=k(x-x0) to find x0 from the given data (x-x0, k and F are given). Then apply it again with F=75.0 N to find x-x0 for the second case.
    Last edited: Nov 17, 2004
  4. Nov 17, 2004 #3
    are the rest right?
  5. Nov 18, 2004 #4


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    There are different ways to solve this one. I'd use an energy approach.
    Since m and v at the beginning are given, you can calculate the kinetic energy [itex]1/2mv^2[/itex].
    The friction does negative work on the sled: W=Fd.
    So you from this you know the kinetic energy at the end. you can get v from this.
    That's correct :smile:
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