This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ouseph (jao888) – HW04 – tsoi – (92940) 1 This printout should have 25 questions. Multiplechoice 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 ....
View
Full
Document
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.
 Summer '08
 Staff
 Physics

Click to edit the document details