Solutions08 - Physics 140 Discussion 08 Solutions 6.20 I...

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Unformatted text preview: Physics 140 - Discussion 08 - Solutions 6.20. I DENTIFY : From the work-energy relation, grav rock . W W K = = ∆ S ET U P : As the rock rises, the gravitational force, , F mg = does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s , the work is negative. Let h be the vertical distance the rock travels. E XECUTE : (a) Applying grav 2 1 W K K =- we obtain 2 2 1 1 2 1 2 2 . mgh mv mv- =- Dividing by m and solving for 1 , v 2 1 2 2 . v v gh = + Substituting 2 15 0 m and 25 0 m/s, h v = . = . 2 2 1 (25 0 m/s) 2(9 80 m/s )(15 0 m) 30 3 m/s v = . + . . = . (b) Solve the same work-energy relation for h . At the maximum height 2 0. v = 2 2 1 1 2 1 2 2 mgh mv mv- =- and 2 2 2 2 1 2 2 (30 3 m/s) (0 0 m/s) 46 8 m. 2 2(9 80 m/s ) v v h g- .- . = = = . . E VALUATE : Note that the weight of 20 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m . Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have an initial velocity of 30.3 m/s. As the rock attains 25....
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Solutions08 - Physics 140 Discussion 08 Solutions 6.20 I...

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