h11_6a - Problem 7.2 In the rough area, the block...

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Unformatted text preview: Problem 7.2 In the rough area, the block decelerates and travels a distance s. We can use the equation v f 2 = v i 2 +2a x. When x =s, v f =0. We can solve for acceleration: a=-v i 2 /2s. We want to find the distance traveled when v f = v i /3. Using v f 2 = v i 2 +2a x, we get x=8s/9 Problem 7.5 We could answer this problem using forces and the equations you learned at the beginning of the class about velocity and acceleration. You might say that since a=gsin , the mass on plane 2 has higher velocity at the bottom. However, the =25 block travels a greater distance to the bottom and has more time to accelerate. Using conservation of energy, mgh=.5mv 2 . Since they start at the same height and the masses are the same, the velocity is the same. Problem 7.6 From conservation of energy, U grav =K+W friction . Energy is dissipated through air friction so W friction >0 and K< U grav Problem 7.8 They have different initial gravitational energies since the masses are different.They have different initial gravitational energies since the masses are different....
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h11_6a - Problem 7.2 In the rough area, the block...

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