A small child is playing with an open-ended cardboard tube that is L = 0.40 mete
ID: 1874308 • Letter: A
Question
A small child is playing with an open-ended cardboard tube that is L = 0.40 meters long, as shown in the figure to the right. He holds the tube at an angle of = 30 and places a small plastic toy with a mass of 0.150 kg through the top end of the tube. Starting from rest, the toy slides down the tube before exiting out the bottom of the tube and falling to the floor. The top end of the tube is h1 = 0.75 meters above the ground and the bottom of the tube is h2 = 0.55 meters above the ground. You may treat the tube as an inclined plane and the coefficient of kinetic friction between the tube and the toy is 0.300.
(a) If the toy is initially at rest at the top of the tube, how fast is the toy traveling when it leaves the tube?
(b) How far does the toy travel horizontally after leaving the tube?
(c) How long does it take the toy to hit the ground once it is released?
Name: Section: uiz 1. A small child is playing with an open-ended cardboard tube that is L-0.40 meters long, as shown in the figure to the right. He holds the tube at an angle of 30° and places a small plastic toy with a mass of 0.150 kg through the top end of the tube. Starting from rest, the toy slides down the tube before exiting out the bottom of the tube and falling to the floor. The top end of the tube is hi -0.75 meters above the ground and the bottom of the tube is h2 0.55 meters above the ground. You may treat the tube as an inclined plane and the coefficient of kinetic friction between the tube and the toy is 0.300 (a) If the toy is initially at rest at the top of the tube, how fast is the toy traveling when it leaves the tube? b) How far does the toy travel horizontally after leaving the tube? h1 c) How long does it take the toy to hit the ground once it is released? h2Explanation / Answer
from figure,
theta = sin^-1((h1-h2)/L)
= sin^-1((0.75 - 0.55)/0.4)
= 30 degrees
a) Net force acting on toy, Fnet = m*g*sin(30) - fk
m*a = m*g*sin(30) - mue_k*N
m*a = m*g*sin(30) - mue_k*m*g*cos(30)
a = g*sin(30) - mue_k*g*cos(30)
= 9.8*sin(30) - 0.3*9.8*cos(30)
= 2.35 m/s^2
now use, vf^2 - vi^2 = 2*a*d
vf = sqrt(2*a*L)
= sqrt(2*2.35*0.4)
= 1.37 m/s
b)
let t is the time taken for the toy to reach the ground.
usem
h2 = voy*t + (1/2)*g*t^2
0.55 = 1.37*sin(30)*t + (1/2)*9.8*t^2
==> t = 0.2723 s
so, horizontal distance travelled, x = vox*t
= 1.37*cos(30)*0.2723
= 0.323 m
c) time taken in the tube, t' = (vf - vi)/a
= (1.37 - 0)/2.35
= 0.583 s
total time taken = t' + t
= 0.583 + 0.272
= 0.855 s