... is found fromsm04.3sm 256 .9kg 050 J)4.231(2J4.23122222v.vmv7. 45: a) J9. 15) m50.2)(sm80.9)(kg 650 .0(2mgh b) The second height is ,m8 75. 1)m50.2( 75. 0 so second ;J9.11mgh ... up the ramp; .sm 25. 6s)m0 .5( m1.6m2 .5 b) In part a), we calculated otherW and 2U. Using Eq. (7.7), J 491.5J 147J 5. 87s)m(11.0kg)12(2212K.sm 05. 9kg)12(J) 5. 491(2222mKv7.7: ... of the ramp as point 2.) In Eq. (7.7), J, 5. 87m)(2.5N) 35( ,0other2 WK and taking 01U and s,m 25. 6J,147)30sin m(2 .5) sm(9.80kg)12(kg12J )5. 87J147(21222vmgyU or sm3.6...