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Homework Help: Mechanics Question: Block on inclined plane

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The situation given in the following question doesn't look possible to me.

    A 1kg block situated on a rough incline is connected to a spring of spring constant 100 N/m. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless. The angle of inclination is 37degree.

    The answer(as given in the book) is 0.125, which can be obtained by the following:
    kx2/2=mgx(sin37-f*cos37) (work-energy theorem)
    where k=100,x=0.1m,m=1kg,f=coefficient of kinetic friction

    2. Relevant equations

    So, mgsin37=1*9.8*0.6=5.88N, and remains constant throughout the descent.
    The spring force increases throughout the distance. At the end of the descent, the force exerted by the spring is kx=100*0.1=10N.
    10>5.88. In addition, it is that 10N is aided by friction.

    So, how can the spring force be greater than mgsin37 and still the body move downward upto that 10cm?

    And the equation of forces when the block is at rest is:
    kx+fmg*cos37=mgsin37
    where f is obviously the coefficient of static friction
    After putting the values, you will get f=-0.52, which is impossible.


    3. The attempt at a solution

    Because of these facts, the situation described in the question doesn't seem probable.
    Is my assessment correct?
     
  2. jcsd
  3. Dec 18, 2009 #2

    ideasrule

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    The pulley??? Where?


    At first, the block accelerates under gravity. At some point it's going to pass the point where gravity is equal to the spring force, but because it has inertia, it keeps on going. Eventually, it comes to a halt and friction keeps it in place.

    The first reason f isn't -0.52 is that friction opposes the spring force, so it should be kx-fmg*cos37=mgsin37. Second, static friction is not equal to N*mu; the maximum possible static friction is. Think about it this way. There's a book resting on your desk. You push it with a tiny force, and friction pushes back with the same force. You push it with a larger force, and friction still pushes back with the same force. If you push it with a force larger than N*mu, friction can't push back with the same force anymore and the book moves.
     
  4. Dec 18, 2009 #3
    Yes. I was silly to ignore that inertia thing. Thanks.
    The pulley: the diagram given shows the spring and pulley on a horizontal surface:
    <Spring>--------<Pulley>
    ----------------------- \ \
    ................................\ \
    ..................................\ <Block>
    ....................................\

    That's how it looks in the diagram.
     
    Last edited: Dec 18, 2009
  5. Dec 18, 2009 #4

    PhanthomJay

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    This equation and statement is not correct. Even though the block is temporarily at rest after its 10 cm descent down the incline, it is still accelerating up the incline, therefore, you cannot use Newton's first law here (you have an unbalanced force acting up the incline). Also, it is kinetic friction we're talking about, not static, since the block is moving down the incline. Although not explicitly stated , the problem is asking for the coefficient of kinetic friction. It is best to use conservation of total energy methods to solve problems of this type (W_friction = delta KE plus delta PE) , or the work energy theorem (W_net = delta KE), although I much prefer the former in this case. (Note that delta KE is zero no matter which of these equations you use).
     
  6. Dec 19, 2009 #5
    @PhantomJay
    Thanks for the post.
    But, what I understood(after getting the reply from ideasrule) is:
    (i)after getting to the end, the body will again move up because of the spring force, unless friction holds it in place
    (ii)I forgot that the unbalanced force will work to decrease the velocity of the block, after spring force becomes more than the downward force.
    (iii)My equation was wrong. The fricition will oppose spring force not the downward one, after reaching the end.
    (iv)If there were no friciton the block would have done periodic motion.
    Yes, while it moves there is kinetic friction. The question demands that. But, the thing I wanted to know was what happens after it has stopped moving. After it stops, the force is from static friction. Correct me if I haven't understood properly.
     
  7. Dec 19, 2009 #6

    PhanthomJay

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    Ok , now you are asking a question different than the original problem question.
    The kinetic coefficient of friction is u_k = 0.125, using energy or other methods. You should convince yourself first that this is correct, by using W_f = delta KE + delta PE. Now when the block comes to a momentary stop at 0.1 m down the plane, examine the forces acting on it: The spring force of kx = 10 N acts up the plane, and the weight component mgsintheta, or about 6 N acts down the plane. The static friction force also acts down the plane (opposing the pending motion up the plane). The static friction coefficient is not given, so let's assume it is equal to the kinetic friction coefficient, u_s =0.125. Then the static friction force at the instant the block comes to a stop is .125(9.8)(.8) = about 1 N. So you have 10N up, and 7 N down, implying a net force of 3 N up, causing an instantaneous acceleration of F_net/m = 3 m/s/s up the plane. The block ultimately slides up and down the plane and settles into its final 'at rest' resting point in its equilibrium position. Does this make sense?
     
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