7KGG4WmZtI4c_1210016121_jwk572

# 7KGG4WmZtI4c_1210016121_jwk572 - oldmidterm 02 KIM, JI Due:...

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Unformatted text preview: oldmidterm 02 KIM, JI Due: Oct 17 2007, noon 1 Question 1, chap 7, sect 2. part 1 of 1 10 points A force vector F = ax + by , where a = 3 . 3 N / m and b = 3 . 1 N / m, acts on an object as the object moves in the direction along the x- axis from the origin to c = 6 . 7 m. Find the work done on the object by the force. Correct answer: 74 . 0685 J (tolerance 1 %). Explanation: The work is found through the expression W = integraldisplay s f s i vector F dvectors. For the force given, W = integraldisplay 6 . 7 m bracketleftBig (3 . 3 N / m) x ( ) + (3 . 1 N / m) y ( ) bracketrightBig dx = integraldisplay 6 . 7 m bracketleftBig (3 . 3 N / m) x ( ) bracketrightBig dx = (3 . 3 N / m) 1 2 x 2 6 . 7 m = (3 . 3 N / m) (22 . 445 m 2 ) = 74 . 0685 J . Question 2, chap 7, sect 3. part 1 of 1 10 points A projectile of mass 0 . 477 kg is shot from a cannon, at height 6 . 3 m, as shown in the figure, with an initial velocity v i having a horizontal component of 6 . 8 m / s. The projectile rises to a maximum height of y above the end of the cannons barrel and strikes the ground a horizontal distance x past the end of the cannons barrel. The acceleration of gravity is 9 . 8 m / s 2 . x v i 4 1 y 6 . 3m Find the magnitude of the final velocity vector when the projectile hits the ground. Correct answer: 14 . 306 m / s (tolerance 1 %). Explanation: Let : v x i = 6 . 8 m / s , x i = 0 m , y i = 6 . 3 m , and = 41 . For the horizontal component of vector v i , we have v x i = v i cos ; consequently v i = v x i cos = (6 . 8 m / s) cos 41 = 9 . 01009 m / s . The work done by gravity depends only on the vertical distance y i , since the work involving y adds and subtracts; i.e. , cancels. The work done by gravity is W = mg y i . From K = 1 2 mv 2 f- 1 2 mv 2 i and the work-energy theorem K = W , the final velocity is 1 2 mv 2 f = W + 1 2 mv 2 i v f = radicalbigg 2 W m + v 2 i = radicalbigg 2 mg y i m + v 2 i oldmidterm 02 KIM, JI Due: Oct 17 2007, noon 2 = radicalBig 2 g y i + v 2 i = bracketleftBig 2 (9 . 8 m / s 2 ) (6 . 3 m) + (9 . 01009 m / s) 2 bracketrightBig 1 / 2 = 14 . 306 m / s . Alternate Solution: The initial velocity in the vertical direction is v y i = v x i tan = (6 . 8 m / s) tan41 = 5 . 91115 m / s , and the vertical velocity v y top = 0 at the top, there- fore we have v 2 y i = v 2 y top + 2 g y , so y = radicalBigg v 2 y i 2 g = radicalBigg (5 . 91115 m / s) 2 2 (9 . 8 m / s 2 ) = 1 . 78274 m . The magnitude of the final vertical velocity v y f is | v y f | = radicalbig 2 g [ y i + y ] = radicalBig 2 (9 . 8 m / s 2 ) [(6 . 3 m) + (1 . 78274 m)] = 12 . 5866 m / s , so v f = radicalBig v 2 y f + v 2 x i = radicalBig (12 . 5866 m / s) 2 + (6 . 8 m / s) 2 = 14 . 306 m / s ....
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## This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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7KGG4WmZtI4c_1210016121_jwk572 - oldmidterm 02 KIM, JI Due:...

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