ch08-p054 - d = 0.292 m being the only positive root(b We...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
54. (a) An appropriate picture (once friction is included) for this problem is Figure 8-3 in the textbook. We apply equation 8-31, Δ E th = f k d , and relate initial kinetic energy K i to the "resting" potential energy U r : K i + U i = f k d + K r + U r ¡ 20.0 J + 0 = f k d + 0 + 1 2 kd 2 where f k = 10.0 N and k = 400 N/m. We solve the equation for d using the quadratic formula or by using the polynomial solver on an appropriate calculator, with
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d = 0.292 m being the only positive root. (b) We apply equation 8-31 again and relate U r to the "second" kinetic energy K s it has at the unstretched position. K r + U r = f k d + K s + U s ¡ 1 2 kd 2 = f k d + K s + 0 Using the result from part (a), this yields K s = 14.2 J....
View Full Document

This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online