07homework06

07homework06 - Husain Zeena – Homework 6 – Due Oct 7...

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Unformatted text preview: Husain, Zeena – Homework 6 – Due: Oct 7 2003, 4:00 am – Inst: H L Berk 1 This print-outshouldhave23questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Fiverampsleadfromthegroundtothesecond floor of a workshop, as sketched below. All five ramps have the same height; ramps B, C, D and E have the same length; ramp A is longer than the other four. You need to push a heavy cart up to the second floor and you may choose any one of the five ramps. Assuming no frictional forces on the cart, which ramp would require you to do the least work? A B C D E 1. Ramp A. 2. Ramp E. 3. Ramp B. 4. Same work for ramps B, C, D or E; less work for ramp A. 5. Ramp C. 6. Ramp D. 7. Same work for the straight ramps A and B; less work for ramps C, D, and E. 8. Same work for all five ramps. correct 9. Unable to determine without knowing the exact profiles of ramps C, D or E. Explanation: Let h be the height of the ramps. Since there is no friction, we can use the work- energy theorem. Work tot = Work person + Work gravity = ΔKE Hence for all the ramps, Work person =- Work gravity + ΔKE = mgh + ΔKE In particular, if Δ KE = 0 (the cart starts from rest and ends at rest), Work person = mgh , for all the ramps. 002 (part 1 of 6) 10 points Given: g = 9 . 8 m / s 2 . The positive x-direction is down the plane. A box of mass 15 . 2 kg with an initial veloc- ity of 1 . 34 m / s slides down a plane, inclined at 26 ◦ with respect to the horizontal. The co- efficient of kinetic friction is 0 . 73. The box stops after sliding a distance x . 1 5 . 2 k g μ k = . 7 3 1 . 3 4 m / s 26 ◦ How far does the box slide? Correct answer: 0 . 420725 m. Explanation: Given : m = 15 . 2 kg , μ k = 0 . 73 , θ = 26 ◦ , and v = 1 . 34 m / s . Basic Concepts: Motion under constant force W = ~ F · ~s P = Δ W Δ t P = dW dt = ~ F · ~v Solution: The net force on the block parallel to the incline is F net = F mg sin θ- F f Husain, Zeena – Homework 6 – Due: Oct 7 2003, 4:00 am – Inst: H L Berk 2 where F f is the friction force. Thus, Newton’s equation for the block reads ma = mg sin θ- F f = mg sin θ- μN = mg (sin θ- μ cos θ ) a = g (sin θ- μ cos θ ) . where N = mg cos θ . To find the distance the block slides down the incline, use v 2 = v 2 + 2 a ( x- x ) , valid for a body moving with a constant accel- eration. Since x = 0 and v f = 0 (the block stops), we get x =- v 2 2 a =- v 2 2 g h sin θ- μ cos θ i = v 2 2 g h μ cos θ- sin θ i = (1 . 34 m / s) 2 2(9 . 8 m / s 2 ) h (0 . 73) cos θ- sin θ i = 0 . 420725 m . 003 (part 2 of 6) 10 points What is the the work done by friction?...
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07homework06 - Husain Zeena – Homework 6 – Due Oct 7...

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