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    -A person weighing 100 N stands on some bathroom scales in a lift. If the scales show a
    reading of 110 N, which answer could describe the motion of the lift?
    Can someone explain why the answer is ""Moving downwards and decelerating.""?


    For Question 1, in which way does tension act and thus, how is it resolved?
    In general, does tension act in the opposite direction to the rope being stretched?

    How do I go about sketching the graph in Question 2 ?

    And for Question 3, Im aware that i have to use gh=(1/2)(v^2) but what height do I take? The mark scheme say's 0.6m and I'm not sure why i must use this height..

    Finally, for Question 4 (part c ) why is the resistive force the horizontal component of force? Shouldn't it be a force that acts in the opposite direction to the moving sled?
    Attached Files
  1. File Type: docx Question 1.docx (144.7 KB, 82 views)
  2. File Type: docx Question 2.docx (136.6 KB, 63 views)
  3. File Type: docx Question 3.docx (157.3 KB, 67 views)
  4. File Type: docx Question 4.docx (405.1 KB, 62 views)
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    (Original post by princejan7)
    -A person weighing 100 N stands on some bathroom scales in a lift. If the scales show a
    reading of 110 N, which answer could describe the motion of the lift?
    Can someone explain why the answer is ""Moving downwards and decelerating.""?
    Let's take up as positive. Then the net force of the person, which is ma is equal to the the normal reaction force of the floor minus weight. (Normal reaction force is equal in magnitude to the reading of the bathroom scales). (1) Try to figure out whether the acceleration is positive (up) or negative (down). (2) In what situations is it possible?


    For Question 1, in which way does tension act and thus, how is it resolved?
    In general, does tension act in the opposite direction to the rope being stretched?
    Yeah, this is more or less correct. Tension is a reaction force. It always acts in the direction of the rope, opposing the force that is exerted on its end tending to stretch it.


    How do I go about sketching the graph in Question 2 ?
    Well, there is no question asked in your attachment. Only a graph is there.


    And for Question 3, Im aware that i have to use gh=(1/2)(v^2) but what height do I take? The mark scheme say's 0.6m and I'm not sure why i must use this height..
    Height h here is the difference in height between the final and initial positions. It's because the change in potential energy is \Delta U=mgh_1-mgh_0=mg\Delta h. And the law of conservation of energy says that \mathrm{(change\ in\ potential\ energy)+(change\ in\ kinetic\ energy)=0}.


    Finally, for Question 4 (part c ) why is the resistive force the horizontal component of force? Shouldn't it be a force that acts in the opposite direction to the moving sled?
    You're right about the frictional force opposing movement. However, in the first sentence it says "The child and sledge are pulled across level ground [...]". Doesn't it imply that they're moving on a horizontal plane?
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    (Original post by jaroc)
    Let's take up as positive. Then the net force of the person, which is ma is equal to the the normal reaction force of the floor minus weight. (Normal reaction force is equal in magnitude to the reading of the bathroom scales). (1) Try to figure out whether the acceleration is positive (up) or negative (down). (2) In what situations is it possible?




    Yeah, this is more or less correct. Tension is a reaction force. It always acts in the direction of the rope, opposing the force that is exerted on its end tending to stretch it.




    Well, there is no question asked in your attachment. Only a graph is there.




    Height h here is the difference in height between the final and initial positions. It's because the change in potential energy is \Delta U=mgh_1-mgh_0=mg\Delta h. And the law of conservation of energy says that \mathrm{(change\ in\ potential\ energy)+(change\ in\ kinetic\ energy)=0}.




    You're right about the frictional force opposing movement. However, in the first sentence it says "The child and sledge are pulled across level ground [...]". Doesn't it imply that they're moving on a horizontal plane?
    Thanks for the reply!
    and I meant for question to 2 : How do I go about sketching the speed time version of the graph?
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    (Original post by princejan7)
    Thanks for the reply!
    and I meant for question to 2 : How do I go about sketching the speed time version of the graph?
    Oh, I see.

    Velocity is the derivative of displacement with regard to time:

    \mathbf{v}=\dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}.

    Therefore if you have a distance-time graph, plot its derivative versus time to obtain the speed-time graph.
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    (Original post by jaroc)
    Oh, I see.

    Velocity is the derivative of displacement with regard to time:

    \mathbf{v}=\dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}.

    Therefore if you have a distance-time graph, plot its derivative versus time to obtain the speed-time graph.
    What if I was not supposed to use calculus?
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    (Original post by princejan7)
    What if I was not supposed to use calculus?

    You wouldn't directly be using calculus. You know that the velocity is the rate of change of displacement. So you would plot the graph of the gradient at each point of the distance/time graph.

    For example at time t = 0, the graph has no gradient so v = 0. The same applies at t = 3.1.

    Does the question give any other information? Such as the car has constant acceleration and constant deceleration? (Because the graph looks like that is the case). If they are constant then when accelerating you'll know that the velocity is increasing at a constant rate - giving a linear relationship (a straight line graph). You'll also know when decelerating the velocity is decreasing at a constant rate - also giving a straight line.


    Edit: Apologies if Jaroc was getting to this, was posted two days ago so I assumed replies would be quite slow.
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    (Original post by Sasukekun)
    Edit: Apologies if Jaroc was getting to this, was posted two days ago so I assumed replies would be quite slow.
    That's great, I wouldn't have explained it better
 
 
 
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