mat-sln-asn-pt6-tst3-sum05

# mat-sln-asn-pt6-tst3-sum05 - Solution for Practice Test 6...

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Unformatted text preview: Solution for Practice Test 6 for Test 3 Solution to Practice Test Problem 6.1() Problem: An airplane is travelling at 100 m s at a height of 200m , and drops a relief package to refugees below, but the parachute attached to the package fails to open. (a)How long does it take for the package to hit ground? (b)How far does the package travel horizontally before it lands? (c)What is the velocity of the package when it lands? Solution to Part (a) y f = y i + v y,i t + 1 2 at 2 Taking y f = 0 , and y i = 200m , and since we know v y,i = 0 , we have Δ y =- 1 2 gt 2 t = radicalBigg- 2Δ y g = radicalBigg- 2(- 200m) 9 . 81 m s 2 = 6 . 4s Grading Key: Part (a) 5 Points 1 point(s) : correct displacement equation for accelerated motion 2 point(s) : no v in direction of acceleration 1 point(s) : correct distance 1 point(s) : correct answer Solution to Part (b) Δ x = x i + v x t = 0 + (100 m s )(6 . 4s) = 640m Grading Key: Part (b) 3 Points 1 point(s) : correct v x 1 point(s) : not accelerated motion 1 point(s) : Δ t from (a) Solution to Part (c) The y-component of the velocity is v y = v y,i + at = 0- gt =- 63 m s The x-component has not changed, so vectorv = 100 m s ˆ x- 63 m s ˆ y Grading Key: Part (c) 3 Points 1 point(s) : v x = v x 1 point(s) : v y = g Δ t 1 point(s) : direction is indicated. Total Points for Problem: 11 Points Solution to Practice Test Problem 6.2(Dragging a suitcase) Problem: A person at an airport is towing her 20kg suitcase at a constant speed by pulling on a strap at an angle θ above the horizontal. She pulls on the strap with a 35N force and the force of friction by the surface on the suitcase is 20N (a)Draw a free body diagram. (b)What angle does the strap make with the horizontal? (c)What is the normal force exerted by the ground on the suitcase? Solution to Part (a) F c ss F c ss,x F c ss,y F n F g es F k fs Grading Key: Part (a) 8 Points 1 point(s) : Force of friction opposite direction of force due to strap 1 point(s) : looks like Σ F x = 0 1 point(s) : looks like Σ F y = 0 1 point(s) : Force due to strap at an angle above horizontal. 1 point(s) : component directions labelled, either here or when start to do (b) 1 point(s) : labels reasonable. 1 point(s) : Force due to gravity present downward 1 point(s) : normal force present upward Solution to Part (b) Start with summing forces: Σ F x = 0 Write the equation for the x-components of the force. The only forces in this direction are the frictional force and the x-component of tension. These are in opposite directions. F k fs- F c ss,x = 0 F k fs = F c ss,x F c ss,x = F c ss cos θ . F k fs = F c ss cos θ Solving, θ = 55 ◦ Grading Key: Part (b) 4 Points 1 point(s) : Realized sum of forces must be zero since speed constant 1 point(s) : tries Σ F x = 0 1 point(s) : correctly takes component of T 1 point(s) : θ = 55 ◦ Solution to Part (c) Now we can write our other force equation. The forces in the y-direction are F c ss,y , the normal force, and the weight of the sled.of the sled....
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mat-sln-asn-pt6-tst3-sum05 - Solution for Practice Test 6...

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