Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can some one tell me how i answer the following questions:

    1)A spring extends by 9cm when a force of 6 N is applied. The limit of proportionality is not exceeded.
    Another identical spring is joined end to end with this spring and a force of 4 N is applied.

    What is the extension of the pair of springs?

    Please somebody tell me how i answer his/

    2) a person weighing 100 N stands on some bathroom scales in a lift. If the scales show a reading of 110 N. What does this describe of the motion of the lift. (e.g moving downwards or upwards, accelerating and declerating??)

    3) The acceleration of free fall on a particular planed is 8.0 ms^-2. An object is dropped from a height and hits the ground after 1.5s. From what height was it dropped??


    Please some one help!!


    Thanks in advance!
    Offline

    2
    ReputationRep:
    1. It's just proportions. 6N to 9cm so what does 4N give?

    2. When the person weighs more than normal the lift is... (try the scenarios with your hands, one hand is person the other the lift floor (it helps me a lot))

    3.  s = ut + 1/2 at^2
    Offline

    0
    ReputationRep:
    (Original post by mujahid_e3)
    Can some one tell me how i answer the following questions:

    1)A spring extends by 9cm when a force of 6 N is applied. The limit of proportionality is not exceeded.
    Another identical spring is joined end to end with this spring and a force of 4 N is applied.

    What is the extension of the pair of springs?

    Please somebody tell me how i answer his/

    2) a person weighing 100 N stands on some bathroom scales in a lift. If the scales show a reading of 110 N. What does this describe of the motion of the lift. (e.g moving downwards or upwards, accelerating and declerating??)

    3) The acceleration of free fall on a particular planed is 8.0 ms^-2. An object is dropped from a height and hits the ground after 1.5s. From what height was it dropped??


    Please some one help!!


    Thanks in advance!
    For number 1 yuse Hookes law F=kx for 3 use SUVAT equations
    Offline

    1
    ReputationRep:
    (Original post by mujahid_e3)
    Can some one tell me how i answer the following questions:

    1)A spring extends by 9cm when a force of 6 N is applied. The limit of proportionality is not exceeded.
    Another identical spring is joined end to end with this spring and a force of 4 N is applied.

    What is the extension of the pair of springs?

    Please somebody tell me how i answer his/

    2) a person weighing 100 N stands on some bathroom scales in a lift. If the scales show a reading of 110 N. What does this describe of the motion of the lift. (e.g moving downwards or upwards, accelerating and declerating??)

    3) The acceleration of free fall on a particular planed is 8.0 ms^-2. An object is dropped from a height and hits the ground after 1.5s. From what height was it dropped??


    Please some one help!!


    Thanks in advance!

    1 - Firstly, find the spring constant...
    Spoiler:
    Show
     F = k \Delta L
    F = Force provided
    k = Spring Constant (constant for BOTH identical springs)
    Delta L = Extension
    Spoiler:
    Show

    k = F / {\Delta L}

= 6/9

= 2/3

    Now;
    {\Delta L} = F / k (for one spring)

= 4 / (2/3)

= 6

    Because there's 2 springs, the answer is an extension of 6cm per spring, so overall, 12cm.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by soup)
    1. It's just proportions. 6N to 9cm so what does 4N give?

    2. When the person weighs more than normal the lift is... (try the scenarios with your hands, one hand is person the other the lift floor (it helps me a lot))

    3.  s = ut + 1/2 at^2
    Can you explain 1 and 2 to me in bit more detail... If you can give the answer i kinda see where your going..

    Thanks for the help soo far
    Offline

    2
    ReputationRep:
    (Original post by mujahid_e3)
    Can you explain 1 and 2 to me in bit more detail... If you can give the answer i kinda see where your going..

    Thanks for the help soo far
    For number 1 do what they've told you up above

    For 2. The initial force is  F = mg where m is mass of person. So that will be 100 = mg. But now its is 110 = mg + k. So what is k? k is the unbalanced force (ma). So we say  110 = mg + ma where a is accleration in lift.

    Now we need to find a situation where there is acceleration. Up is +ve, down is -ve (since acceleration is a vector)

    So accelerating upwards is +ve (accelerating is +ve) times +ve (direction) which is +ve. It's going to give a +ve answer which would be added on to 100. This may explain the increase from 100 to 110
    Accelerating downwads is +ve times -ve whice is -ve (direction) . This would be taken away from 100.
    Decellerating upwards is -ve (decellerating is -ve) times +ve whice is -ve. This would be taken away from 100.
    Decellerating downwards is -ve times -ve which is +ve. It's going to give a +ve answer which would be added on to 100. This may explain the increase from 100 to 110
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by soup)
    For number 1 do what they've told you up above

    For 2. The initial force is  F = mg where m is mass of person. So that will be 100 = mg. But now its is 110 = mg + k. So what is k? k is the unbalanced force (ma). So we say  110 = mg + ma where a is accleration in lift.

    Now we need to find a situation where there is acceleration. Up is +ve, down is -ve (since acceleration is a vector)

    So accelerating upwards is +ve (accelerating is +ve) times +ve (direction) which is +ve. It's going to give a +ve answer which would be added on to 100. This may explain the increase from 100 to 110
    Accelerating downwads is +ve times -ve whice is -ve (direction) . This would be taken away from 100.
    Decellerating upwards is -ve (decellerating is -ve) times +ve whice is -ve. This would be taken away from 100.
    Decellerating downwards is -ve times -ve which is +ve. It's going to give a +ve answer which would be added on to 100. This may explain the increase from 100 to 110


    so for 1 do you just find the K? if not what do i do?
    Offline

    2
    ReputationRep:
    (Original post by mujahid_e3)
    so for 1 do you just find the K? if not what do i do?
    Yes just do what Ian. said
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by soup)
    1. It's just proportions. 6N to 9cm so what does 4N give?

    2. When the person weighs more than normal the lift is... (try the scenarios with your hands, one hand is person the other the lift floor (it helps me a lot))

    3.  s = ut + 1/2 at^2
    Sorry to keep bothering you, but how would you actually do q3? Because you dont have u , you only have a and t? So please explain

    Thanks
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Ian.)
    1 - Firstly, find the spring constant...
    Spoiler:
    Show
     F = k \Delta L
    F = Force provided
    k = Spring Constant (constant for BOTH identical springs)
    Delta L = Extension
    Spoiler:
    Show

    k = F / {\Delta L}

= 6/9

= 2/3

    Now;
    {\Delta L} = F / k (for one spring)

= 4 / (2/3)

= 6

    Because there's 2 springs, the answer is an extension of 6cm per spring, so overall, 12cm.
    Thanks for that.. You helped alot.. Could you explain q3?? please

    Thanks
    Offline

    2
    ReputationRep:
    (Original post by mujahid_e3)
    Sorry to keep bothering you, but how would you actually do q3? Because you dont have u , you only have a and t? So please explain

    Thanks
    The object was dropped so the initial velocity was zero
    Offline

    1
    ReputationRep:
    (Original post by mujahid_e3)
    Sorry to keep bothering you, but how would you actually do q3? Because you dont have u , you only have a and t? So please explain

    Thanks
    u = 0
    The object was dropped.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks everybody!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.