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Applied maths Q4 help

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  • 05-05-2014 5:13pm
    #1
    Registered Users Posts: 27


    Any idea how to do part ii ? I'm not sure how to find the acceleration.


Comments

  • Closed Accounts Posts: 609 ✭✭✭thirteen.


    Whats the question?


  • Registered Users Posts: 27 c0unterpart


    thirteen. wrote: »
    Whats the question?

    I cant add pictures could I dm the question or will i have to type it out


  • Closed Accounts Posts: 609 ✭✭✭thirteen.


    Upload the picture to an image hosting site and link it.


  • Registered Users Posts: 27 c0unterpart


    http:// tinypic.com/view.php?pic=24w9fky&s=8#.U2e7C3y9KSM

    no space


  • Closed Accounts Posts: 609 ✭✭✭thirteen.


    The acceleration of the 7kg block will be the same as the 3kg block, and it will stay like that until it falls off the edge of the table because no external forces are acting upon it.


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  • Registered Users Posts: 27 c0unterpart


    thirteen. wrote: »
    The acceleration of the 7kg block will be the same as the 3kg block, and it will stay like that until it falls off the edge of the table because no external forces are acting upon it.

    Thanks.

    Am i going about finding the speed right by saying 3(v) = 7(u) where i have the correct value for v ?


  • Closed Accounts Posts: 609 ✭✭✭thirteen.


    I don't know what it is you are trying to do there tbh. but you are definitely over complicating it.

    Its really quite simple.

    a=whatever, s=6.615, t7=?
    s=1/2at^2 to get t7, then take t3 from t7 to get how much longer it takes to reach the edge of the table.


  • Registered Users Posts: 137 ✭✭Nicke011


    I agree with thirteen.
    7kg is there just to trick you, because the table is smooth so there is no friction, therefore the acceleration of the bodies are the same.
    I got 0.82 seconds for (I) and 0.34 seconds for (II) (t=1.16-0.82=0.34s)


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