i)
"W=Fd=12\\cdot3=36" J
F - horizontal force, d - distance
ii)
kinetic energy:
"E=W=36" J
iii)
"E=mv^2\/2"
"v=\\sqrt{2E\/m}=\\sqrt{2\\cdot36\/200}=0.6" m/s
iv)
"ma=F"
"a=F\/m=12\/200=0.06" m/s2
"v=at+v_0"
"d=at^2\/2+v_0t+d_0"
if start velocity "v_0=0" and start distance "d_0=0" , then:
"t=\\sqrt{2d\/a}"
final velocity:
"v=a\\sqrt{2d\/a}=0.06\\sqrt{2\\cdot3\/0.06}=0.6" m/s
Comments
Leave a comment