Answer to Question #260237 in Physics for Janine

Question #260237

a 100 lb rests on the floor and the coefficient of sliding friction between block and floor is 0.25. a horizontal force of 40 lb acts on the block for 3s accelerate it. compute the velocity of the block at the end of 3s.


1
Expert's answer
2021-11-03T10:22:20-0400

Let's first find the acceleration of the block:


FhFfr=ma,F_h-F_{fr}=ma,FhμsW=Wga,F_h-\mu_sW=\dfrac{W}{g}a,a=FhμsWWg,a=\dfrac{F_h-\mu_sW}{\dfrac{W}{g}},a=40 lb0.25×100 lb100 lb32.17 fts2=4.82 fts2,a=\dfrac{40\ lb-0.25\times100\ lb}{\dfrac{100\ lb}{32.17\ \dfrac{ft}{s^2}}}=4.82\ \dfrac{ft}{s^2},a=4.82 fts2×0.3048 m1 ft=1.47 ms2.a=4.82\ \dfrac{ft}{s^2}\times\dfrac{0.3048\ m}{1\ ft}=1.47\ \dfrac{m}{s^2}.

Finally, we can find the velocity of the block at the end of 3s from the kinematic equation:


v=v0+at=0+1.47 ms2×3 s=4.41 ms.v=v_0+at=0+1.47\ \dfrac{m}{s^2}\times3\ s=4.41\ \dfrac{m}{s}.

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