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A 90 N block rests on a horizontal surface. It takes 40 N to get it moving and 25 N to keep it moving. Calculate the coefficients of static and kinetic friction.

Physics

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Answer

Static friction coefficient

0.44\text{0.44}

Kinetic friction coefficient

0.28\text{0.28}

Step-by-step solution

4 steps

Here's the solution to question 2:

  1. Identify the given values.

    The weight of the block is W=90 NW = 90 \text{ N}.
    The force required to get the block moving (static friction) is Fs=40 NF_s = 40 \text{ N}.
    The force required to keep the block moving at a constant velocity (kinetic friction) is Fk=25 NF_k = 25 \text{ N}.

  2. Calculate the normal force.

    For a block on a horizontal surface, the normal force FNF_N is equal to its weight WW.

    FN=W=90 NF_N = W = 90 \text{ N}
  3. Calculate the coefficient of static friction (μs\mu_s).

    The formula for static friction is Fs=μsFNF_s = \mu_s F_N.
    Rearrange to solve for μs\mu_s:

    μs=FsFN\mu_s = \frac{F_s}{F_N}

    Substitute the values:

    μs=40 N90 N\mu_s = \frac{40 \text{ N}}{90 \text{ N}} μs≈0.44\mu_s \approx 0.44
  4. Calculate the coefficient of kinetic friction (μk\mu_k).

    The formula for kinetic friction is Fk=μkFNF_k = \mu_k F_N.
    Rearrange to solve for μk\mu_k:

    μk=FkFN\mu_k = \frac{F_k}{F_N}

    Substitute the values:

    μk=25 N90 N\mu_k = \frac{25 \text{ N}}{90 \text{ N}} μk≈0.28\mu_k \approx 0.28

    The coefficients of friction are:
    Static friction coefficient: 0.44\boxed{\text{0.44}}
    Kinetic friction coefficient: 0.28\boxed{\text{0.28}}

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