Question

N1 weight Work and Kinetic Energy - 4 Forces - Horizontal Motion This problem is similar to lecture example and discussion problem. As shown in the figure, a 25.5 kg block is pulled to the right by force F = 79 N. The coefficient of kinetic friction between the block and the floor is uk = 0.26. The block moves through a distance of 18.00 m. DON"T use scientific notation. Review on application of Newton's second law. Step 1: Start by drawing a free body diagram of the block, including all the forces exerted on the block. Label the normal force as FN, label the weight as weight, the kinetic frictional force fk. You should be able to calculate the weight of the block. Force F is given. But the other 2 forces: the frictional force fk and the normal force FN are unknown. For each force, write its x component and y component. (Use FN as "space holder", write fk with its formula). Step 2: Apply Newton's second law Apply F = may, what should ay be equal to?, obtain an equation. (a) Solve for the magnitude of the normal force FN (in Newtons), it is: Enter a number N Submit (5 attempts remaining)

Answer 1

a) m = 25.5kg Fiin 40 f=79N MK = 0.26 F10140 S=18m mg a) { Fy=may >> FN + FBingo-mg.emio DEN amg -f Sim 40 = 725.5X9-81-798m 40 FN = 199137N7 bieden WCF) =Ě= F31080 WCFN) = FNS 8: (08 90 = 0 7 b) ft = khoe hc 'd be a = He N= 0. 2 6 X 14 9:37 = 5108419 w ( fr ) = fr. § COB (180) --S]+84 x 8 == {331 | 2 c) WF) = F S 10$40 = 79x1810840 = 1.089.322 do wlng) = mg.s 10890 = 07 Network done zw - WIFN 14 W ( fr 2 +WIFI+W/ng) -0-933.12+1089-32+0 = 156.27

E f> Network = change ink E= 2m (v²vo?) 156-2=12x 25.5 (V² 2:22) > 13222 = 28156.2 25.5 2x 156.2 N 25.5 +2.22 = 4.13 mis speed = V = 4:13 mis