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

Version 001 – Energy Practice – Ross – (20691) 76. W = N Dcosθ7. W = −N D8. W = N DExplanation:The normal force makes an angle of 90 ◦with the displacement, so the work done by itis zero.023 (part 3 of 3) 0.0 pointsWhat is the final speed of the block?1. v =√2m (F cosθ −mgsinθ −µ kN)DcorrectThusv f =√2m (F cosθ −mgsinθ −µ kN)D.Cannon 01024 (part 1 of 6) 0.0 pointsA projectile of mass 0.314 kg is shot froma cannon, at height 6.7 m, as shown in thefigure, with an initial velocity v i having ahorizontal component of 6.4 m/s.Theprojectilerisestoamaximumheightof∆y above the end of the cannon’s barrel andstrikes the ground a horizontal distance ∆xpast the end of the cannon’s barrel.v i∆y2. v =√2m(F cosθ +mgsinθ)D3. v =√2m (F sinθ −µ kN)D51 ◦6.7 m4. v =√2m (F cosθ −mgsinθ +µ kN)D5. v =√2m (F cosθ +mgsinθ −µ kN)D6. v =√2m(F cosθ −mgsinθ)D7. v =√2m (F cosθ −µ kN)D8. v =√2m (F sinθ +µ kN)DExplanation:The work done by gravity isW grav = mgDcos(90 ◦ +θ)= −mgDsinθ.The work done by the force F isW F = F Dcosθ.From the work-energy theorem we know thatW net = ∆K,W F +W grav +W friction = 1 2 mv2 f .∆xDetermine the vertical component of theinitialvelocity at the end of the cannon’s barrel,where the projectile begins its trajectory.The acceleration of gravity is 9.8 m/s 2 .Correct answer: 7.90334 m/s.Explanation:Let : v xi = 6.4 m/s andθ = 51 ◦ .tanθ = v yiv xiv yi = v xi tanθ = (6.4 m/s)tan51 ◦= 7.90334 m/s .025 (part 2 of 6) 0.0 pointsDetermine the maximum height ∆y the projectileachieves after leaving the end of thecannon’s barrel.