1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newton's Laws problems

  1. Sep 28, 2006 #1
    1) A 5kg box is pulled by a force F horizontally (no angle). There is friction between the box and horizontal surface. Mu_k = .25 and Mu_s = .5. If F=20N find the magnitude of the friction force and the acceleration of the block.

    I used the formula |f_k| = mu_k * |Normal force|

    f_k = .25 * 9.8 * 5
    f_k = 12.25 N

    The correct answer is that the friction is 20 N, meaning that the block didn't move. Where did I go wrong?

    2) A 5kg sled is pulled an an angle 5 degree above the horizontal. Friction between the sled and the snow is described by mu_k = .1 and mu_s = .15. What is acceleration of the sled?

    The correct answer is 2.99 m/s^2

    This is what I did.

    I took the 20 N and multiply by cos (5 degree) to get 19.92.

    |f_k| = mu_k * N
    f_k = .1*5*9.8
    f_k = 4.9

    (19.92 - 4.9) / 5 = a
    a = 3.004

    Did I just round my answer differently or did I do something wrong?

    3) A 2kg object is under the influences of several forces. Assume that the net force is constant. At t=0, the object's velocity is 3i + 2j and at t=4s its velocity is 11i - 14j . Find the magnitude and direction of net force acting on the object. Please explain this to me. I have no idea how to start.

    Thanks in advance for your help.
  2. jcsd
  3. Sep 28, 2006 #2
    Can you please tell me what Mu_k, Mu_s and |f_k| all mean? I'm doing similar stuff to this at the moment but I don't recognise those things you are using.
  4. Sep 28, 2006 #3
    Mu_k is the coefficient of kinetic friction
    Mu_s is coefficient of static friction

    |f_k| is kinetic friction
    |f_s| is static friction
  5. Sep 28, 2006 #4
    Oh okay gotcha. Yeah I actually probably can't help because I haven't yet delved into static and kinetic friction really. I have worked with simply friction before without determining differences between the two
  6. Sep 28, 2006 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Before the box moves, it is static friction that is important. You calculated the acceleration using kinetic friction. Go back and do it using static friction.
  7. Sep 28, 2006 #6
    Thanks for your help. Since the static friction is greater than the force pushing the object, the object will not move. Since 20N is applied on the object, the friction is also 20N according to Newton's Third Law. Am I correct?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Newton's Laws problems