el work-energy theorem equation with what does the box reach the bottom roller c
ID: 1608774 • Letter: E
Question
Explanation / Answer
5.a)
conserve energy
KE = PE
0.5mv^2 + mgh= mgH
0.5v^2 = 9.81(30-20)
v = 14.00 m/s
b.)energy lost = total enegy at A- total enegy at C
E1 = mgH - mgh-0.5mv^2
E =m(30-20) - 0.5m*2*2
E1 = 96.1m
E =mgH = m*9.81*30 = 294.3m
so percentage of energy lost = (96.1m/294.3m)100 = 32.65%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.