HW_07_Solutions_F05 - Physics 112 - HW #7 Solutions 6.4...

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- 1 - Physics 112 - HW #7 Solutions Fall 2005 6.4 [Sliding Crate] (a) Crate is sliding at constant speed. Apply Newton’s 2 nd Law: x direction Σ F x = F a – f k r = ma = 0 F a = f k r (1) y direction Σ F y = N - mg = 0 (2) Kinetic friction: f k r = µ k N Substitute from (1) and (2): F a = µ k mg = (0.25)(30.0 kg)(9.81 m/s 2 ) = 73.6 N (b) Work done on crate by applied force F a : W = F a .s = (73.6 N)(4.5 m) = 331 J (c) Work done on crate by friction f k : W = f k .s = (-73.6 N)(4.5 m) = -331 J (d) Work done on crate by normal force and by gravity are both 0 J , since these forces are both perpendicular to the displacement. (e) Total work done on crate is 0 J . We know this because crate does not accelerate, i.e. its KE does not change. 6.13 [Alpine Rescue] Work done on box by friction to travel uphill along slope by a distance l : W f = f k .s = - f k l (1) This work is negative because the friction force is in the opposite direction to the displacement. Increase in vertical height of box is h : l sin α = h, or l = h/sin α (2) Applying Newton’s law in direction perpendicular to
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW_07_Solutions_F05 - Physics 112 - HW #7 Solutions 6.4...

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