HS5 - Adams Russell – Homework 5 – Due midnight –...

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Unformatted text preview: Adams, Russell – Homework 5 – Due: Oct 12 2006, midnight – Inst: Opyrchal, H 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A block is released from rest on an inclined plane and moves 3 . 4 m during the next 4 s. The acceleration of gravity is 9 . 8 m / s 2 . 1 5 k g μ k 34 ◦ What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 622198 . Explanation: Given : m = 15 kg , ‘ = 3 . 4 m , θ = 34 ◦ , and t = 4 s . Consider the free body diagram for the block m g s i n θ N = m g c o s θ μ N a mg The acceleration can be obtained through kinematics. Since v = 0, ‘ = v t + 1 2 at 2 = 1 2 at 2 a = 2 ‘ t 2 (1) Applying Newton’s Second Law of Motion X F i = ma and Eq. 1, the sine component of the weight acts down the plane and friction acts up the plane. The block slides down the plane, so ma = mg sin θ- μ k mg cos θ 2 ‘ t 2 = g ‡ sin θ- μ k cos θ · 2 ‘ = g t 2 ‡ sin θ- μ k cos θ · μ k = g t 2 sin θ- 2 ‘ g t 2 cos θ (2) = tan θ- 2 ‘ g t 2 cos θ = tan34 ◦- 2(3 . 4 m) (9 . 8 m / s 2 )(4 s) 2 cos34 ◦ = . 622198 . keywords: 002 (part 1 of 2) 10 points A car is traveling at 47 . 3 mi / h on a horizontal highway. The acceleration of gravity is 9 . 8 m / s 2 . If the coefficient of friction between road and tires on a rainy day is 0 . 082, what is the minimum distance in which the car will stop? (1 mi = 1.609 km) Correct answer: 278 . 074 m. Explanation: Newton’s second law in the direction of motion gives- f k =- μ k mg = ma Solving for a a =- μ k g (1) The acceleration a may be found from the following kinematics equation with v f = 0: v 2 f = v 2 + 2 ax a =- v 2 2 x (2) Combining equations (1) and (2), we obtain v 2 2 x = μ k g Adams, Russell – Homework 5 – Due: Oct 12 2006, midnight – Inst: Opyrchal, H 2 or solving for x , x = v 2 2 μ k g Plugging in the appropriate values, x wet = (21 . 1405 m / s) 2 2(0 . 082)(9 . 8 m / s 2 ) = 278 . 074 m 003 (part 2 of 2) 10 points What is the stopping distance when the sur- face is dry and μ dry = 0 . 536? Correct answer: 42 . 5411 m. Explanation: x dry = (21 . 1405 m / s) 2 2(0 . 536)(9 . 8 m / s 2 ) = 42 . 5411 m (A considerable difference) keywords: 004 (part 1 of 1) 10 points Masses 19 kg and 11 kg are connected by a light string that passes over a frictionless pulley as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 19 kg 11 kg μ If the 19 kg mass, initially held at rest on the table, is released and moves 1 . 7 m in 1 . 5 s, determine the coefficient of kinetic friction between it and the table....
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HS5 - Adams Russell – Homework 5 – Due midnight –...

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