Answer in Mechanics | Relativity for Mwansa Kunda #126533
5 kg body locates 4 meters above ground. After it released it will go down because it heavier than 3 kg body. Energy conservation principle says:
“E_{p}=E_{k}”
(1) “mtimes atimes h = frac{mtimes V^2}{2}”
where m – mass of body (5 kg), a – acceleration (not equals to gravitational acceleration in our case because we have 3kg body on other side), h – height (4 meters), V – velocity of body near ground.
acceleration could be found:
“F_{result} = F_{g5} – F_{g3}= m_{5}times g – m_{3} times g;”
“F_{result} = 5times 9.8 – 3times 9.8 = 49 – 29.4 = 19.6N”;
“F_{result}=m_{5}times a;”
“a=frac{F_{result}}{m_{}}=frac{19.6}{5}=3.92 m/s^2”
so, returning to formula (1) we can find V of body 5kg and body 3 kg (they a the same in this moment)
“V=sqrt{frac{m_{5}times a times h times 2}{m_{5}}}=sqrt{a times h times 2}=sqrt{3.92times 4times 2};”
“V=5.6m/s”
Ansver for first question speed is 5.6 m/s
After 5 kg body hit ground, body 3kg will locate on height 8 m and continue move upward for some time, because it has some speed. To find maximum height which body 3 will have, let’s use the same principle:
“m_{3}times gtimes h=frac{m_{3}times V^2}{2};”
“h=frac{V^2}{2times g}”
this time we use g because only force that affects body 3 kg is gravity.
“h=frac{5.6^2}{2times 9.8}=1.6m”
So maximal height of body 3 kg is 8 + 1.6 = 9.6m
