10_03_2nd_law_Energy

10_03_2nd_law_Energy - f equal 0 So use ∆ x = v ave ∆ t v ave =(v i v f/2 v f = v i ∆ v and ∆ v = a ∆ t Hint After two algebraic

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Physics 211 10/1 HW Handout Due Monday 10/7 Fall 2002 2 nd Law vs. Energy 1) Paul pushes a 10kg box across a rough floor. The box’s initial velocity is 0.5m/s to the right. The coefficicient of kinetic friction is μ k = 0.3 Paul applies a steady force of 50N to the right on the box. Let g = 10N/kg for this problem. a) Draw a free body diagram at right consistent with the second law, including arrows showing the directions of the acceleration and the net force. b) Find the acceleration, magnitude and direction, of the box. Show your reasoning. c) Find the velocity when the box has moved a distance of 5m. You will need to find the time but neither v i or v
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Unformatted text preview: f equal 0. So, use ∆ x = v ave ∆ t, v ave = (v i + v f )/2, v f = v i + ∆ v, and ∆ v = a ∆ t. Hint: After two algebraic substitutions you get ∆ x = v i ∆ t + 1 / 2 a ∆ t 2 . Write this as 1 / 2 a ∆ t 2 +v i ∆ t - ∆ x = 0, plug in the numbers for a and v i then solve the quadratic for ∆ t. Use the quadratic formula that says if a t 2 + b t + c = 0 Then: t = (-b ± √ b 2 – 4 ac )/2 a where a = 1 / 2 a, b = v i, c = -∆ x (Don’t forget to find the final velocity) d) Now find the same final velocity using energy buckets and “work.” Show your reasoning. a = F net = box rough floor Paul...
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This note was uploaded on 12/21/2010 for the course PHYS Physics 21 taught by Professor Larrywatson during the Fall '08 term at North Dakota.

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