A factory worker pushes a 31.7 crate a distance of 5.0 along a level floor at co

Question

A factory worker pushes a 31.7 crate a distance of 5.0 along a level floor at constant velocity by pushing downward at an angle of 30 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.

Part A: What magnitude of force must the worker apply to move the crate at constant velocity?

Part B:How much work is done on the crate by this force when the crate is pushed a distance of 5.0 ?

Part C:How much work is done on the crate by friction during this displacement?

Part D:How much work is done by the normal force?

Part E:How much work is done by gravity?

Part F:What is the total work done on the crate?

Explanation / Answer

Mass of the crate m = 31.7 Kg Acceleration a = 0 Coefficient of kinetic friction = 0.26 Let F be the applied force. Distance covered S = 5 m Friction f = [mg + F sin30] (A) From Newtons second law we can write                          ma = F cos30 - f                            0 = 0.866 F - 0.26[ 310.66 + F sin30]                 0.866 F = 80.77 + 0.13 F                  0.736 F = 80.77                               F = 109.74 N (B) Work done by the aplied force W1 = FS cos30                                                       = 109.74 * 5 *0.866                                                       = 475.189 J (C) Work done by the friction W2 = f S cos180                                                 = - f *S                                                 = - [mg + F sin30]S                                                 = -0.26[310.66 + 54.87]5                                                 = - 475.189 J (D) Work done by the normal force W3 = NS cos90 = 0                                                    (E) Work done by the gravity W4 = mg S cos90 = 0 (F) Net work done W = W1 + W2 + W3 + W4                              = 475.189 - 475.189 + 0 + 0                              = 0 J (C) Work done by the friction W2 = f S cos180                                                 = - f *S                                                 = - [mg + F sin30]S                                                 = -0.26[310.66 + 54.87]5                                                 = - 475.189 J (D) Work done by the normal force W3 = NS cos90 = 0                                                    (D) Work done by the normal force W3 = NS cos90 = 0 (E) Work done by the gravity W4 = mg S cos90 = 0 (F) Net work done W = W1 + W2 + W3 + W4                              = 475.189 - 475.189 + 0 + 0                              = 0 J

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