P207 - Fall 2007 Solutions Assignment 8 1. Chapter 13, Problem 37 (a) The escape velocity v = 2GM/R. The gravitational acceleration at the surface is g = GM/R2 . Therefore, v = 2Rg = 2(500 103 m)(3 m/s2 ) = 1732 m/s. (b) The energy of a particle at
P207 - Fall 2008 Solutions Assignment 5
1 (a) (b) The sum of the forces in the horizontal direction Fx = max . The acceleration is 1 v2 ax = , (1) 2d where v is the initial velocity and d the distance over which the veloc
P207 - Fall 2008 Solutions Assignment 6 1. Chapter 7, Problem 16 Since F = ma, we know that the displacement of the canister will be in the same direction as the net force acting on it. Therefore W = |F |d The net force is F = F1 + F2 + F3 i = (3.0 N
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Prelim 2 Exam
November 11, 2003
PLEASE CHECK THAT THIS PACKAGE HAS 11 PROBLEMS ON 11 PAGES!
Three blank pages to be used as scrap paper should be attached at the end of
A ball of mass m is swung at the end of the rope in a
horizontal circle. Its speed v is constant and the length of
the rope is L.
What work W is done by the tension T in the rope when
the mass moves a small distance s along the circle?
A. mg s
B. T s
P207 - Fall 2007 Solutions Assignment 6 1. Chapter 7, Problem 13 (a) The work done by the forces is x Fx + y Fy and we know that y = 0 so the forces in the y-direction make no contribution. In the x-direction Fx = |F2 | cos - |F1 | = (9 N) cos 60 -
P207 - Fall 2007 Solutions Assignment 7 1. Chapter 8, Problem 28 (a) The total mechanical energy of the box is the sum of the potential energy of the spring, the potential energy of gravity and the kinetic energy. Let x = 0 when the spring is in the
P207 - Fall 2007 Solutions Assignment 10 1. Chapter 10, Problem 48 O = FA RA sin A + FB RB sin B + FC RC sin C = (10 N)(8 m) sin(135 ) - (16 N)(4 m) sin(90 ) + (19 N)(3 m) sin(160 ) = 12 Nm (Counter Clockwise) 2. Chapter 12, Question 5. (a) The verti