A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.

Question

A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.5 m and h2=2 m by an ideal spring as shown in Fig. 2. The spring constant is k=22. 103 N/m. The first part of the path is frictionless but on the horizontal part after the hill there is friction that can be described by a kinetic coefficient of friction µk=0.4.

a) The brick is launched with this minimal spring compression. Calculate how far along the horizontal friction track the brick moves before coming to rest.

Chrome File Edit View History Bookmarks People Window Help ] a 99% E Sun Oct 25 3:06:48 PM Rama Chaiyya Chaiyya (Er https://blackboard.. CA Red Brick with M × C Two Bricks With Ma X Your question need × C A Red Brick With M Homeworks- AS.17 X × https://blackboard.hu.edu/bbcswebdav/pid-2809400 t content d 11532627 2/courses AS 171 103.01 FA15/BlackBoard%20Assignment%20796. n energy during this collision and the associated velocities of the bricks after the collision. (c) After the collision m2 actually ends up moving a velocity v2f-2 m/s. How much mechanical energy was converted into thermal energy during the collision? Problem 3 A red brick with mass m=4.5 kg is launched over terrain defined by heights h1=2.5 m and h-2 m by an ideal spring as shown in Fig. 2. The spring constant is k-2210 N/m. The first part of the path is frictionless but on the horizontal part after the hill there is friction that can be described by a kinetic coefficient of friction =0.4 (a) Calculate the smallest compression of the spring that will bring the brick over the hill (b) The brick is launched with this minimal spring compression. Calculate how far along the horizontal friction track the brick moves before coming to rest Figure 2 9 Assignment 1 1-doc AFC Transportation docx- 20151017_152554-1 (1).jpg 11-20151017-153 149-1(1).jpg -1-20151017_153203 (1).jpg Show All x 25 BOK

Explanation / Answer

1) using energy conservation,

initial energy = kx^2/2

final energy = mgh1

kx^2 /2 = mgh1

22 * 10^3 * x^2 /2 = 4.5 * 9.81 * 2.5

x = 0.10 m

2) again using energy conservation to find speed at the botttom of the hill,

mg(h1 +h2) = m v^2 /2

v = sqrt(2 x 9.81 x (2.5 + 2) ) = 9.40 m/s

and after that friction will act in opposite direction.

friction = uk N = uk mg

a = - uk g = - 0.4 x 9.81 m/s^2

using v^2 - v^2 = 2ad

0 - (9.40^2 )= 2 x -0.4 x 9.81 x d

d= 11.25 m

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