AS Physics - Mechs questions

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    Any help with any of these much appreciated

    1. A person is standing on the floor of a lift which is accelerating uniformly upwards.
    a) Draw a free body diagram showing all the forces acting upon the person
    b) What are the Newton's Third Law reactions to each of these and on what does each act?
    c) Which, if any, of the forces in a) change as the acceleration changes? Give brief reasons.

    2a) A tug is towing two barges at constant speed as shown in attachment. The tension in cable CD is 12000N. Calculate the tension in each of the cables AC and BC.
    b) The speed of the tug is 1.8ms^-1 and 25% of the power of the propulsion system is used to drive the tug and barges. Calculate the power of the propulsion system.

    3. Can do most of this q, about stopping distances, apart from one part:
    Suggest a PHYSICAL explanation for the shape of this graph (braking distance against speed, braking distance proportional to speed squared). I assume it is looking for something other than KE = 1/2mv^2, but not sure of the "physical" reason for this.

    4. A transport plane is towing a glider, which in turn is towing a second glider. They accelerate from rest on a level runway. The plane and gliders remain in a straight line and the tow ropes are horizontal. The gliders each have a mass of 1000kg. Assume that the force opposing the motion of each glider is constant at 1500N. The plane reaches a speed of 40ms^-1 after accelerating uniformly over a distance of 320m.
    b)During the acceleration of the plane (I calculated this in a) as 2.5ms^-1?), calculate
    i) the tension in the tow rope between the two gliders
    ii) the tension in the tow rope between the plane and the first glider
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    a)you simply drow a man with the normal acting upwords contact forcce from the left ground and the weight acting downwords
    b)every action has an opposing reaction same value and opposite direction or you can say
    when body A applies a force on body B body B applies and equal opposite force on body A
    the pictue in a is in equalibrium but if r changed the acceleration changes you can changes the weight unless you changed the person or the person suddenly lost wieght (mass)
    dont feel like readsing the rest right now,I'll try to answer them laster if nobody did
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    (Original post by habosh)
    for 1
    a)you simply drow a man with the normal acting upwords contact forcce from the left ground and the weight acting downwords
    b)every action has an opposing reaction same value and opposite direction or you can say
    when body A applies a force on body B body B applies and equal opposite force on body A
    the pictue in a is in equalibrium but if r changed the acceleration changes you can changes the weight unless you changed the person or the person suddenly lost wieght (mass)
    dont feel like readsing the rest right now,I'll try to answer them laster if nobody did
    Thanks.. but since part b) asks for the reaction, would it not just be mg for part a), as is the contact force not the reaction and would that not therefore go in part b)? But I thought that that was wrong because it says forceS...
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    (Original post by dinkymints)
    Thanks.. but since part b) asks for the reaction, would it not just be mg for part a), as is the contact force not the reaction and would that not therefore go in part b)? But I thought that that was wrong because it says forceS...
    the lift exterts an equal oppsite force on the dude,which is called contact force ,as the weight exterts force on the lift downwords the man exterts force on the lift,if there were no upword force on the man they man would go in the lift *dig downwords* ,which is impossible...so R or what is called the normal reaction is pretty important
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    I think part b) wants you to say the reaction to the weight is the force of the boy pulling on the earth (acting on the earth) and the reaction to the contact force is the force of the boy pushing on the lift (acting on the lift).
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    for 2)

    Apply newton II horizontally (to the right)

    12000 - xcos25 - xcos25 = 0 ----(x=tension in each of other rope)

    xcos25=6000

    x = 6620N

    b) Power = Fv = 12000 x 1.8 = 21600W but this is only 25% of power supplied,
    so power supplied = 21600 / 0.25 = 86400W
 
 
 
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