Version 001 – Energy Practice – Ross – (20691) 1 This print-out ...

Version 001 – Energy Practice – Ross – (20691) 3FF59 ◦θm382 NF µmF hHow much force is exerted by the ropeon the crate? The acceleration of gravityis 9.8 m/s 2 .Correct answer: 741.693 N.Explanation:Let : F h = 382 N andθ = 59 ◦ .Thehorizontalcomponentoftheforceexertedby the rope is defined bycosθ = F hFF = F hcosθ = 382 Ncos59 ◦ = 741.693 N .009 (part 2 of 3) 0.0 pointsWhat work is done by Sally if the crate ismoved 76.2 m?Correct answer: 29108.4 J.Explanation:The motion is in the direction of the horizontalcomponent, so the work Sally does isW = F h d = (382 N)(76.2 m)= 29108.4 J .010 (part 3 of 3) 0.0 pointsWhat work is done by the floor through forceof friction between the floor and the crate?Correct answer: −29108.4 J.Explanation:Since there is no acceleration, all of herwork is done against friction, so the frictionforce is −W.Holt SF 05Rev 10011 (part 1 of 3) 0.0 pointsA flight attendant pulls her 72.4 N flight baga distance of 289 m along a level airport floorat a constant velocity. The force she exerts is39.3 N at an angle of 52.6 ◦ above the horizontal.a) Findthe work she does on theflight bag.Correct answer: 6898.39 J.Explanation:Basic Concept:Given:Solution:W = F dcosθF g = 72.4 NF applied = 39.3 Nd = 289 mθ = 52.6 ◦W 1 = F k ·d= F applied d cosθ= (39.3 N)(289 m) cos52.6 ◦= 6898.39 J.012 (part 2 of 3) 0.0 pointsb) Find the work done by the force of frictionon the flight bag.