{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 5 - homework 04 JACOBS AMBER Due 1:00 am Question 1 chap 6...

This preview shows pages 1–3. Sign up to view the full content.

homework 04 – JACOBS, AMBER – Due: Feb 19 2008, 1:00 am 1 Question 1, chap 6, sect 2. part 1 of 2 10 points A block is released from rest on an inclined plane and moves 3 . 4 m during the next 5 . 2 s. The acceleration of gravity is 9 . 8 m / s 2 . 12 kg μ k 34 What is the magnitude of the acceleration of the block? Correct answer: 0 . 251479 m / s 2 (tolerance ± 1 %). Explanation: Given : m = 12 kg , = 3 . 4 m , θ = 34 , and t = 5 . 2 s . Consider the free body diagram for the block m g sin θ N = m g cos θ μ N a m g The acceleration can be obtained through kinematics. Since v 0 = 0, = v 0 t + 1 2 a t 2 = 1 2 a t 2 a = 2 t 2 (1) = 2 (3 . 4 m) (5 . 2 s) 2 = 0 . 251479 m / s 2 . Question 2, chap 6, sect 2. part 2 of 2 10 points What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 643556 (tolerance ± 1 %). Explanation: Applying Newton’s Second Law of Motion summationdisplay F i = m a 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 m a = m g sin θ - μ k m g cos θ 2 t 2 = g parenleftBig sin θ - μ k cos θ parenrightBig 2 = g t 2 parenleftBig sin θ - μ k cos θ parenrightBig μ k = g t 2 sin θ - 2 g t 2 cos θ (2) = tan θ - 2 g t 2 cos θ = tan 34 - 2 (3 . 4 m) (9 . 8 m / s 2 ) (5 . 2 s) 2 cos 34 = 0 . 643556 . Question 3, chap 6, sect 5. part 1 of 2 10 points Given: The coefficient of static friction be- tween the person and the wall is 0 . 43 , the mass of the person is 59 kg , the radius of the cylinder is 8 . 3 m , and g = 9 . 8 m / s 2 . A barrel of fun consists of a large vertical cylinder that spins about the vertical axis. When it spins fast enough, any person inside will be held up against the wall.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
homework 04 – JACOBS, AMBER – Due: Feb 19 2008, 1:00 am 2 ω 8 . 3 m Find ω c , the critical angular speed below which a person will slide down the wall of the cylinder. Correct answer: 1 . 65707 rad / s (tolerance ± 1 %). Explanation: Let : m = 59 kg , R = 8 . 3 m , and μ = 0 . 43 . The normal force on the person by the wall provides the centripetal acceleration N = m ω 2 R . The maximum force of static friction is given by f = μ N . In equilibrium we have f = m g . Hence m g = μ m ω 2 c R . or solving for ω c ω c = radicalbigg g μ R = radicalBigg 9 . 8 m / s 2 (0 . 43) (8 . 3 m) = 1 . 65707 rad / s . Question 4, chap 6, sect 5. part 2 of 2 10 points Compare the following two cases where for each case the angular speed is greater than the critical value as defined in Part 1. For the first case, denote its angular speed and frictional force by ω 1 and f 1 , respectively. For the second case, its angular speed is ω 2 = 2 ω 1 . Express f 2 , the frictional force for the sec- ond case, in terms of f 1 . 1. f 2 = f 1 correct 2. f 2 = 3 f 1 3. f 2 = f 1 3 . 5 4. f 2 = f 1 2 5. f 2 = 2 f 1 6. f 2 = 4 f 1 7. f 2 = 1 . 5 f 1 8. f 2 = f 1 3 9. f 2 = f 1 4 10. f 2 = 2 . 5 f 1 Explanation: Since for each case the angular speed is above the critical angular speed, the person will remain in equilibrium in the vertical di- rection. Hence f 2 = f 1 = m g .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}