Lil asked in Science & MathematicsPhysics · 1 month ago

What is the weight of this block?

A weight W is now placed on the block and

7.7 N is needed to push them both at constant

velocity.

What is the weight W of the block?

Answer in units of N.

Update:

So there was a part one to this problem. Here it is. Sorry about the confusion.

A block weighing 9.4 N requires a force of 2 Nto push it along at constant velocity.What is the coefficient of friction for thesurface? 

3 Answers

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  • NCS
    Lv 7
    1 month ago
    Favorite Answer

    "constant velocity" means "equilibrium," and so the applied horizontal force must equal the friction force.

    So from the initial problem, we found from

    friction force = µ * normal force

    that

    µ = 2N / 9.4N = 0.21

    and so for the current problem

    friction force = µ * new normal force

    7.7 N = 0.21 * (W + 9.4N)

    which solves to

    W = 27 N

    as the answers should have no more than 2 significant digits (one might argue for 1 significant digit owing to the data "2 N.")

    Hope this helps!

  • oubaas
    Lv 7
    1 month ago

    a)

    μ = F/W = 2/9.4 = 0.213

    b)

    (W+W1)*0.213 = 7.7

    W1 = (7.7-9.4*0.213)/0.213 = 26.75 N

  • 1 month ago

    Please provide the rest of the problem.

    • Lil1 month agoReport

      Never mind I found the rest of the info.

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