Ch. 8 Solutions

# Ch. 8 Solutions - Chapter 8 Problem 2. Problem Solutions...

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

Chapter 8 Problem Solutions Problem 2. Now take Fig. 8-26 to lie in a vertical plane, and find the work done by the gravitational force as an object moves from point 1 to point 2 over each of the paths shown. Solution Take the origin at point 1 in Fig. 8-26 with the x -axis horizontal to the right and the y -axis vertical upward. The gravitational force on an object is constant, F g = ! mg ˆ j , while the paths are (a) d r = ˆ j dy for x = 0 and 0 ! y ! l , followed by d r = î dx for y = l and 0 ! x ! l , and (b) d r = î dx + ˆ j dy = ( î + ˆ j ) dy , for 0 ! y ! l (since x = y along this path). The work done by gravity (Equation 7-11) is W g ( a ) = F g ! d r = " ( # mg ˆ j ) ! ˆ j dy 0 \$ " + ( # mg ˆ j ) ! î dx 0 \$ " = # mg dy + 0 = # mg l , 0 \$ " and W g ( b ) = ( ! mg ˆ j ) " ! ( î î + ˆ j 0 l # ) dy = ! mg dy 0 l # = ! mg l . Of course, these must be the same because gravity is a conservative force. Problem 10. A 60-kg hiker ascending 1250-m-high Camel’s Hump mountain in Vermont has potential energy ! 2.4 " 10 5 J; the zero of potential energy is taken at the mountain top. What is her altitude? Solution If we measure the altitude from sea level, U ( y ) ! U ( 1250 m ) = mg ( y ! 1250 m ) = ! 2 . 4 " 10 5 J , or y = 1250 m ! 2 . 4 ! 10 5 J = (60 kg) ( 9 . 8 m/s 2 ) = 842 m . Problem 10 Solution. Problem 17. A particle moves along the x -axis under the influence of a force F = ax 2 + b , where a and b are constants. Find its potential energy as a function of position, taking U = 0 at x = 0. Solution Equation 8-2a, with U (0) = 0, gives U ( x ) = ! F x d " x = ! 0 x # ( a " x 2 + b ) d " x 0 x # = ! 1 3 ax 3 ! bx .

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

View Full Document