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    A block of wood of mass 2kg is pushed across a rough horizontal floor. The block moves at 3m/s and the coefficient of friction is 0.55. Calculate the work done in 2 sec.

    So far I know that the friction force is μR = 0.55*2*9.8=10.78N.
    Also the distance travelled s=ut so 3m/s*2s=5m. I got no clue of how to use this information to find the actual force to work out the energy. I was thinking of doing f=ma so like
    Σf=ma
    F-10.78=ma...

    I'm so stuck


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    (Original post by bigboateng)
    A block of wood of mass 2kg is pushed across a rough horizontal floor. The block moves at 3m/s and the coefficient of friction is 0.55. Calculate the work done in 2 sec.

    So far I know that the friction force is μR = 0.55*2*9.8=10.78N.
    Also the distance travelled s=ut so 3m/s*2s=5m. I got no clue of how to use this information to find the actual force to work out the energy. I was thinking of doing f=ma so like
    Σf=ma
    F-10.78=ma...

    I'm so stuck


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    3 times 2=6 NOT 5.
    Work done against friction=friction force times distance moved.
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    (Original post by brianeverit)
    3 times 2=6 NOT 5.
    OMG lol. So i can C1-4. d1,d2, fp1-2, M1 and I can't multiply 3*2 😂😂😂

    But thank you correcting it to 6 works now.


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