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Numerical · Q38

Q.A body of mass 37 kg rests on a rough horizontal surface. The minimum horizontal force required to just start the motion is 68.5 N. In order to keep the body moving with constant velocity, a force of 43 N is needed. What is the value of a) coefficient of static friction? and b) coefficient of kinetic friction?

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✓ Free question

Step 1. Given: mass m=37m = 37 kg, force to just start motion FL=68.5F_L = 68.5 N, force to sustain constant velocity Fk=43F_k = 43 N.

Step 2. Normal reaction: N=mg=37×9.8=362.6N = mg = 37\times9.8 = 362.6 N.

Step 3. (a) μs=FLN=68.5362.6=0.1889≈0.188\mu_s = \dfrac{F_L}{N} = \dfrac{68.5}{362.6} = 0.1889 \approx 0.188.

Step 4. (b) μk=FkN=43362.6=0.1186≈0.119\mu_k = \dfrac{F_k}{N} = \dfrac{43}{362.6} = 0.1186 \approx 0.119 (the book prints 0.118; the difference is due to rounding at the third decimal place — both values are correct to the given data).

✓Final answer

a) μs = 0.188 b) μk = 0.118 (μk ≈ 0.119 by direct unrounded calculation — the two are the same to two significant figures)

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