{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 5-solutions

# HW 5-solutions - kuruvila(lk5992 HW 5 opyrchal(11113 This...

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

kuruvila (lk5992) – HW 5 – opyrchal – (11113) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 76 N block rests on a table. The sus- pended mass has a weight of 36 N. 76 N 36 N μ s What frictional force is required to keep the blocks from moving? Correct answer: 36 N. Explanation: Given : W 1 = 76 N and W 2 = 36 N . a T N W 1 μ s N a T W 2 Consider the forces then the blocks would first start to move: For block W 2 vertically, F y,net = T - W 2 = 0 For block W 1 horizontally, F x,net = T - μ N = 0 μ N = T = W 1 = 36 N . 002 (part 1 of 2) 10.0 points A car is traveling at 54 . 1 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 . 077, what is the minimum distance in which the car will stop? (1 mi = 1 . 609) Correct answer: 387 . 396 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 0 + 2 ax a = - v 2 0 2 x (2) Combining equations (1) and (2), we obtain v 2 0 2 x = μ k g or solving for x , x = v 2 0 2 μ k g Plugging in the appropriate values, x wet = (24 . 1797 m / s) 2 2(0 . 077) (9 . 8 m / s 2 ) = 387 . 396 m 003 (part 2 of 2) 10.0 points What is the stopping distance when the sur- face is dry and μ dry = 0 . 645? Correct answer: 46 . 2473 m. Explanation: x dry = (24 . 1797 m / s) 2 2(0 . 645) (9 . 8 m / s 2 ) = 46 . 2473 m (A considerable difference)

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

View Full Document
kuruvila (lk5992) – HW 5 – opyrchal – (11113) 2 004 10.0 points The magnitude of each force is 208 N, the
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern