# 09exam2 - Husain Zeena – Exam 2 – Due 9:00 pm – Inst...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Husain, Zeena – Exam 2 – Due: Oct 15 2003, 9:00 pm – Inst: H L Berk 1 Thisprint-outshouldhave14questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A 5 . 5 kg mass on a frictionless inclined surface is connected to a 2 . 4 kg mass. The pulley is massless and frictionless, and the connecting string is massless and does not stretch. The 2 . 4 kg mass is acted upon by an upward force of 4 . 1 N, and has a downward acceleration of 4 . 7 m / s 2 . 5 . 5 k g μ = θ 2 . 4 kg 4 . 1 N 4 . 7m / s 2 What is the tension in the connecting string? Correct answer: 8 . 14 N. Explanation: Given : m 1 = 5 . 5 kg , m 2 = 2 . 4 kg , F = 4 . 1 N , and a = 4 . 7 m / s 2 . Basic Concept: X i ~ F i = m~ a Solution: It is easiest to analyze the forces on each block separately. m 2 F m 2 g T a m 1 a T m 1 g s i n θ N m 1 g c o s θ Note: T is the tension in the string, F is the external force, and the weight of m 2 is m 2 g . From the free-body diagram on the 2 . 4 kg block m 2 a = m 2 g- F- T , so T = m 2 g- m 2 a- F = (2 . 4 kg)(9 . 8 m / s 2- 4 . 7 m / s 2 )- 4 . 1 N = 8 . 14 N . 002 (part 1 of 1) 10 points Given: The surface is frictionless, Two blocks in contact with each other move to the right across a horizontal surface by the two forces shown. 1 . 9 kg 5 . 8 kg 76 N 13 N μ = 0 Determine the magnitude of the force ex- erted on the block with mass 1 . 9 kg by the block with mass 5 . 8 kg. Correct answer: 60 . 4545 N. Explanation: Given : F 1 = 76 N , F 2 = 13 N , M 1 = 1 . 9 kg , M 2 = 5 . 8 kg , and μ = 0 . Husain, Zeena – Exam 2 – Due: Oct 15 2003, 9:00 pm – Inst: H L Berk 2 M 1 M 2 F 1 F 2 μ = 0 a The common acceleration of the system is F 1- F 2 = ( M 1 + M 2 ) a a = F 1- F 2 M 1 + M 2 . (1) This result is obtained by treating the two masses as a single object. M 1 + M 2 F 2 F 1 ( M 1 + M 2 ) g N To determine the force exerted on M 1 by M 2 , we need to analyze M 1 separately. M 1 F 1 F 12 M 1 g N From the free body diagram of M 1 we see that the only two horizontal forces acting on M 1 are the external force F 1 and the force F 12 exerted by M 2 , resulting in the acceleration of the mass F 1- F 12 = M 1 a, so F 12 = F 1- M 1 a F 12 = F 1- M 1 • F 1- F 2 M 1 + M 2 ‚ F 12 = (76 N)- (1 . 9 kg) × • (76 N)- (13 N) (1 . 9 kg) + (5 . 8 kg) ‚ F 12 = (76 N)- (15 . 5455 N) = 60 . 4545 N . 003 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A curve of radius r is banked at angle θ so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right....
View Full Document

## This note was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 7

09exam2 - Husain Zeena – Exam 2 – Due 9:00 pm – Inst...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online