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Numerical · Q24

Q.A block of mass 5 kg5\ \text{kg} rests on a horizontal floor for which the coefficient of static friction is 0.50.5 and the coefficient of kinetic friction is 0.40.4. Find

(a) the minimum horizontal force needed to just start the block moving, and
(b) the horizontal force needed to keep it moving at constant velocity once it has started. (Take g=9.8 m/s2g = 9.8\ \text{m/s}^2.)
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Given: mass m=5 kgm = 5\ \text{kg}, μs=0.5\mu_s = 0.5, μk=0.4\mu_k = 0.4, g=9.8 m/s2g = 9.8\ \text{m/s}^2. Normal reaction on the horizontal floor: N=mg=5×9.8=49 NN = mg = 5 \times 9.8 = 49\ \text{N}.\n\n**(a) To just start motion**, the applied force must reach the maximum static friction: Fstart=μsN=0.5×49=24.5 NF_{\text{start}} = \mu_s N = 0.5 \times 49 = 24.5\ \text{N}\n\n**(b) To keep it moving at constant velocity** once sliding, the applied force need only balance the (smaller) kinetic friction: $$F_{\text{constant}} = \mu_k N = …

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