Recent content by Monic

  1. M

    Calculating Force for an Exceptional Vertical Jump

    Finally it all makes sense! :D thank you a lot to everyone, as for conservation of energy I just don't know how to apply it to this problem as it uses joules not force but that's fine
  2. M

    Determining max horizontal force using friction coefficient

    The maxiumum value of force is the same as the Ff (0.175 as shown above) If I calculate accelertion 0.175N/0.5kg=a i get 0.35m/s^2 If I put that acceleration to calculate the max force of the other block (1kg)(0.35m/s2)=F i get 0.35N again...which is wrong
  3. M

    Calculating Force for an Exceptional Vertical Jump

    okay all I did was v^2=2gh v=2(9.8)(0.8) v=3.95 the other velocity is 0 then I put 0^2=3.95^2 +2a(0.2) to find the acceleration needed to find the force for f=ma a=39 f=39(70kg) f=2730N which is still wrong, unless added the normal force he needs to overcome would be the correct thing to...
  4. M

    Calculating Force for an Exceptional Vertical Jump

    okay so the initial velocity is 3.96 m/s and v2 is 0m/s at the top of the jump, therefore 0^2=3.96m/s^2 +2(a)0.8 giving a= 9.8...which is wrong FInding the constant force that needs to be applied at 0.2m to get this person that high must be greater than the normal force f=mg which is...
  5. M

    Calculating Force for an Exceptional Vertical Jump

    I am aware of conservation of energy and tried it but it did not seem to work, I must be making an error. I do not know how to find the initial velocity... would v^2=2(9.8)(0.8m) be correct? v=3.96m/s as v? I am really lost on this one :( thank you for the welcome :)
  6. M

    Calculating Force for an Exceptional Vertical Jump

    Homework Statement An exceptional vertical jump from rest would raise a person 0.8m off the ground. To do this, what constant force would a 70kg person have to exert against the ground? Assume that the person lowers himself by 0.2m prior to jumping and remains in a standing position while in...
  7. M

    Determining max horizontal force using friction coefficient

    Homework Statement A 0.5 kg wooden block is placed on top of a 1.0kg block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.2. What is the maximum horizontal force that can be applied to the...
  8. M

    Stopping distance using velocity and friction coefficient on an incline plane

    Thank you so much! That gets the exact answer in the book (75m) Much appreciated :)
  9. M

    Stopping distance using velocity and friction coefficient on an incline plane

    Homework Statement A skier skiing downhill reaches the bottom of a hollow with a velocity of 20m/s, and then coasts up a hill with a 10° slope. If the coefficient of kinetic friction is 0.10, how far up the slope will she travel before she stops? Homework Equations aΔt=v2-v1 d= 1/2(v1 +...
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