Lesson_3.2_Homework_Solution

Lesson_3.2_Homework_Solution - Lesson 3.2 Homework...

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Lesson 3.2 Homework Solutions 1. A 2.0-kg box is raised to a height of 3.0 m by applying a vertical force of 50 N. Find (a) the work done by the applied force. (b) the work done by gravity (c) the gain in kinetic energy of the box. (d) the gain in potential energy of the box. (a) Work done by the applied force = 50 N × 3 m = 150 J (b) Work done by gravity = -mgh = -2 × 9.8 × 3 = -58.8 J (c) The gain in kinetic energy = net work done = 150 – 58.8 = 91.2 J (d) The gain in potential energy is the negative of the work done by the force of gravity. U = -(-58.8 J) = 58.8 J Alternatively: [] 23 3 10 0 3 0 29 . 8 1 9 . 58.8 J 6 19.6 19.6(3 0) h h um g d y d y d y y ∆=−− =−−× = = == = ∫∫ 2. A box of mass 2.0 kg initially moving at 5.0 ms -1 is acted upon by a constant force of 25 N in the direction of motion. After some time the speed of the box is 22 ms -1 . Calculate the work done by the force on the box. Initial kinetic energy of the box = (1/2)mv o 2 = (1/2) × 2 × 5 2 = 25 J Final kinetic energy of the box = (1/2)mv f 2 = (1/2) × 2 × 22 2 = 484 J W = E = 484 – 25 = 459 J 3. A spring gun has a spring with a force constant of 500 Nm -1 . If this gun is used to shoot a 25-g ball straight up by compressing the spring by 6 cm, find: (a) The initial kinetic energy of the ball. (b) The potential and kinetic energies at the highest point. (c) The height reached by the ball. Potential energy of the compressed spring is:
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U = (1/2)kx 2 = (1/2) × 500 × 0.06 2 = 0.9 J (a) This potential energy of the spring is converted to the initial kinetic energy of the ball. Therefore, initial kinetic energy of the ball = 0.9 J (b) At the highest point all the energy is potential = 0.9 J (c) mgh = 0.9 0.025 × 9.8 h = 0.9 h = 3.67 m 4.
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This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_3.2_Homework_Solution - Lesson 3.2 Homework...

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