With μk= 0.25, Eq. 6-2 leads to fk= 11 N. (b) We apply Newton’s second law to the xaxis: ()215 N cos 4011 Ncos0.14 m/s3.5 kgkFfmaaθ°−−=¡==.Since the result is positive-valued, then the block is accelerating in the +x(rightward) direction.7. We choose +xhorizontally rightwards and +
This is the end of the preview. Sign up
access the rest of the 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.