Tarzan swings on a 24.0 m long vine initially inclined at an angle of 35° from the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he starts with an initial speed of 2.78 m/s?
2007-11-25 05:16:28 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Use conservation of energy
PE=sin(35)*24*g*m
initial KEi=.5*m*vi^2
KE at the bottom
KEo=.5*m*vo^2
a) vi=0
sin(35)*24*m*g=.5*m*vo^2
vo=sqrt(sin(35)*48*g)
b vi=2.78
sin(35)*24*m*g+
.5*m*2.78^2=.5*m*vo^2
vo=sqrt(sin(35)*48*g+2.78^2)
j
2007-11-25 05:22:41 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
the change in height of tarzan w.r.t ground will give us the change in potential energy..
change in height = 24 (1-cos 35)
= 4.34
a) 1/2*m*v^2 = m*g*4.34
--> v = 9.2 m/s
b) 1/2*m*v^2 = m*g*4.34 + 1/2*m*u^2
1/2*v^2 = 42.53 + 3.86
--> v= 9.63 m/s
2007-11-25 13:27:41 · answer #2 · answered by The learner 2 · 0⤊ 0⤋
lol i had the same exact problem to do. This is a classic centripetal force problem so use the centripetal force equations.
2007-11-25 13:24:39 · answer #3 · answered by Dmitriy 2 · 0⤊ 0⤋