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# solution_4 - ouseph (jao888) – HW04 – tsoi – (92940)...

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Unformatted text preview: ouseph (jao888) – HW04 – tsoi – (92940) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A block is released from rest on an inclined plane and moves 2 . 5 m during the next 3 . 4 s. The acceleration of gravity is 9 . 8 m / s 2 . 1 5 k g μ k 29 ◦ What is the magnitude of the acceleration of the block? Correct answer: 0 . 432526 m / s 2 . Explanation: Given : m = 15 kg , ℓ = 2 . 5 m , θ = 29 ◦ , and t = 3 . 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, ℓ = v t + 1 2 a t 2 = 1 2 a t 2 a = 2 ℓ t 2 (1) = 2 (2 . 5 m) (3 . 4 s) 2 = . 432526 m / s 2 . 002 (part 2 of 2) 10.0 points What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 503847. Explanation: Applying Newton’s Second Law of Motion summationdisplay 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 ℓ t 2 = g parenleftBig sin θ − μ k cos θ parenrightBig 2 ℓ = g t 2 parenleftBig sin θ − μ k cos θ parenrightBig μ k = g t 2 sin θ − 2 ℓ g t 2 cos θ (2) = tan θ − 2 ℓ g t 2 cos θ = tan29 ◦ − 2 (2 . 5 m) (9 . 8 m / s 2 ) (3 . 4 s) 2 cos 29 ◦ = . 503847 . 003 (part 1 of 3) 10.0 points The suspended 2 . 6 kg mass on the right is moving up, the 2 kg mass slides down the ramp, and the suspended 7 . 7 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 11 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. ouseph (jao888) – HW04 – tsoi – (92940) 2 2 k g μ = . 1 1 36 ◦ 7 . 7 kg 2 . 6 kg What is the acceleration of the three block system? Correct answer: 4 . 85824 m / s 2 . Explanation: Let : m 1 = 2 . 6 kg , m 2 = 2 kg , m 3 = 7 . 7 kg , and θ = 36 ◦ . Basic Concept: F net = ma negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di- rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo- nent of its weight is mg sin θ and the perpen- dicular component of its weight is mg cos θ . ( N = mg cos θ from equilibrium). The accel- eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μm 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ − T 1 − μm 2 g cos θ ....
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## This note was uploaded on 07/09/2011 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.

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solution_4 - ouseph (jao888) – HW04 – tsoi – (92940)...

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