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A Few Questions (Can Someone Check My Work?)

  1. Oct 15, 2005 #1
    The following link shows the problems I have. I have solved them (hopefully); I just need some confirmation or someone to tell me where I went wrong. I'd appreciate it.


    1. I am going to say the answer is choice #2, mgsinθ up incline. This one seemed pretty obvious so I'm not going to write out my work unless someone thinks I am incorrect.

    2. It didn't say that the banana is resting on anything, so I am going to say F3 only is the reaction force to F1 (answer choice #2).

    3. At the highest point, the upward force is equal in magnitude to the downward force (0,-mg), so they cancel out meaning the weight is not involved. However, at the highest point, the baseball only experiences an x velocity, which is not a force (since there isn't accel.). Because of this, I am going to say answer choice #4 (no forces on the ball). However, it is difficult the way the question is worded. I could also say that there is a downward force, it is just matched (cancelled out) by the upward force. ??

    4. Since the wedge is frictionless, the mass doesn't effect acceleration. m1 obviously experiences more accel. due to gravity because the angle is greater, so I am going to say choice #1 (m1 reaches the bottom first).

    5. I figured the net force to be 20 N to the right. Now I divided by the total mass 3kg+2kg and came up with an acceleration of 4m/s2. I figured the force that m exerts of M is m*4=2*4=8N. Is this right or should I have multiplied the acceleration by the mass M instead?

    I would really appreciate it if someone would comment on these. Thank you very much for your time and help :smile:

    Attached Files:

  2. jcsd
  3. Oct 15, 2005 #2


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    Yes, that's correct.
    Yes, that's correct also.
    Where did you get the idea that "At the highest point, the upward force is equal in magnitude to the downward force"?
    Once the ball has left the bat there is no upward force! Neglecting air resistance, there is only one force acting on the ball at any point in its flight.
    Yes, since they have to go the same distance, the one with the greater acceleration will reach the bottom first and that is m1.
    You are correct that the net force is 20 N to the right and that the two masses move together with acceleration 4 m/s2.
    Now, what net force on M (3 kg mass) would produce an acceleration of 4 m/s2? Since F1 is 20 N, what other force is necessary to give that net force?
    Another way to do this problem is to look at m. What force on m (2 kg mass) would produce an acceleration of 4 m/s2 to the right? Since F2 is 10 N to the left, what other force is necessary to give that net force?
  4. Oct 15, 2005 #3
    -10 N + F3 = 8 N --> F3 = 18 N

    Is this right? Thanks.
  5. Oct 17, 2005 #4
    I was just looking at the question with the banana (#2) and realized that "none of the above" would be the correct answer because there's only one body involved. Is this right?

    Thanks again.
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