A ball of mass 2.08 kg is tied to a string of length 2.34 m as shown in Figure P
ID: 2250768 • Letter: A
Question
A ball of mass 2.08 kg is tied to a string of length 2.34 m as shown in Figure P6.50. The ball is initially hanging vertically and is given an initial velocity of 7.6 m/s in the horizontal direction. The ball then follows a circular arc as determined by the string. What is the speed of the ball when the string makes an angle of 27° with the vertical?
A ball of mass 2.08 kg is tied to a string of length 2.34 m as shown in Figure P6.50. The ball is initially hanging vertically and is given an initial velocity of 7.6 m/s in the horizontal direction. The ball then follows a circular arc as determined by the string. What is the speed of the ball when the string makes an angle of 27Â degree with the vertical?Explanation / Answer
initial kinetic energy of the ball = 0.5* m*v*v
given m = 2.08 kg
v= 7.6 m/s
joules
initial KE = 0.5*2.08*7.6*7.6= 60.0704
given string of length 2.34 m
height raised when the string makes an angle 27 degrees
= L-Lcos 27
=2.34(1-cos 27)
=0.255 meters
increase in potential energy = mgh = 2.08*9.81*0.255 = 5.204 joules
using conservation of energy
we have final kinetic energy = initial KE-increase in PE
= 60.0704 - 5.204
=54.8664
equating with 0.5*2.058*v*v
we have v*v = 54.8664/(0.5*2.058)
v = 7.302 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.