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

Version 001 – Energy Practice – Ross – (20691) 4Correct answer: −6898.39 J.Explanation:Basic Concept: Because the bag’s speedis constant,Solution:F net = F k +F applied cosθ = 0W 2 = −W 1= −F k ·d= −F applied dcosθ= −(39.3 N)(289 m)cos52.6 ◦= −6898.39 J.013 (part 3 of 3) 0.0 pointsc) Find the coefficient of kinetic friction betweenthe flight bag and the floor.Correct answer: 0.579654.Explanation:Basic Concept:Solution:µ k = F kF nF k = µ k F n = µ k F g .= −F appliedcosθ−F g +F applied sinθ−(39.3 N)cos52.6 ◦=−(72.4 N)+(39.3 N)sin52.6 ◦−(39.3 N)cos52.6◦=−(41.1795 N)= 0.579654.Holt SF 05Rev 60014 (part 1 of 3) 0.0 pointsA 5.0 kg block is pushed 3.0 m at a constantvelocity up a vertical wall by a constant forceapplied at an angle of 30.0 ◦ with the horizontal,as shown in the figure.The acceleration of gravity is 9.81 m/s 2 .5 kgF3 m30 ◦Drawing not to scale.If the coefficient of kinetic friction betweenthe block and the wall is 0.20, finda) the work done by the force on the block.Correct answer: 225.141 J.Explanation:Basic Concepts:W F = F y dcosθ ′ = Fdsinθsince θ ′ = 0 ◦ ⇒ cosθ ′ = 1.Because v is constant,Given:Solution:F y,net = F sinθ−µ k F n −mg = 0F x,net = F cosθ −F n = 0F n = F cosθm = 5.0 kgθ = 30.0 ◦d = 3.0 mµ k = 0.20g = 9.81 m/s 2F sinθ −µ k F n = mgF sinθ −µ k F cosθ = mgmgF =sinθ −µ k cosθW F = Fdsinθmgdsinθ=sinθ−µ k cosθ= (5 kg)(9.81 m/s2 )sin30 ◦ (3 m)sin30 ◦ −(0.2)cos30 ◦= 225.141 J