Question

Work and Kinetic Energy - 4 Forces - Horizontal Motion This problem is similar to lecture example and discussion problem. As shown in the figure, a 23 kg block is pulled to the right by force F = 76 N. The coefficient of kinetic friction between the block and the floor is uk = 0.27. The block moves through a distance of 15.00 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 Fare unknown. For each force, write its x component and y component. (Use F as "space holder", write fk with its formula). Step 2: Apply Newton's second law Apply Fy - may, what should ay be equal to obtain an equation. (a) Solve for the magnitude of the normal force FN (in Newtons), it is: 176.55 N Correct (100.0%) Submit Before calculating work, DRAW a DISPLACEMENT VECTOR on the fgure. Calculate the work (in Joules) done by the normal force. Pay attention to the angle in the work formula and the sign of work. Incorrect (0.09) (Correct answer: 0.00) Submit (b) What is the magnitude (in Newtons) of the kinetic frictional force? Enter a number N Submit (5 attempts remaining) Calculate the work (in Joules) done by the kinetic frictional force. Pay attention to the angle in the work formula and the sign of work. Enter a number Submit 15 attempts remaining)

Answer 1

Quer Up = 0.27 . fsinuo F = 760 m= 23kg S= 15m g = 9.8 m/s² so y po srcu Tan 17 Fcos uo = FBD 4 For Normal Force N= mg - F sinuo° = (23) 09.8) - 76 8in 40° - N =176.5 3) Work Done by Normal = 0=goo so [w=o (6) fiction force = ein = (0.27) (17605) =47.6 Newton 21 cfecüction = -fs = 647-6)(15) = -715 J © Whore = (Fcos 40°) (15) = 873. 2 I do wng = mg (5) cos 90º = zero 1 le total = o-757873.2 to = 158.2 Newton