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CATAGORY: Mechanics
QUESTION: A 2kg block is released with an initial speed of 5 m/s along
a horizontal tabletop toward a light spring with force
constant 10 N/m. The coefficient of Kinetic friction between
block and table is 0.25, and the block is initially 2.0 m
from the spring. a) What is the max compression of spring?
b) How far from the spring is the block when it comes to rest?
```

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```
The initial kinetic energy of the block is

1         1
(1) KE =  - m v^2 = - 2kg (5m/s)^2 = 25 J
2         2

This energy is dissipated by friction AND goes into the potential
energy of the spring.  The potential energy of the spring is

1         1
(2) PE = - k x^2 = - 10 N/m x^2 = 5.0 N/m x^2
2         2

Where x is the compression length. the energy dissipated by friction is

(3) EF = Uk mg (2m + x) = .25 2kg 9.8m/s^2(2m + x)

where Uk is the coeficient of friction, m is the mass if the block and
g = 9.8 m/s^2.  We include a 2m + x because while the spring is
compressing the block is still sliding.  Most solution sets omit this
part because x is usualy small compared to the sliding length but we
include it out of formality. From conservation of all energy we have

(4) KE = PE + EF

Substitution of (1), (2), and (3) into (4) gives

(5) 25 J = 5.0 N/m x^2 + 4.9(2m + x)

or

(6)  0 = 5.0 N/m x^2 + 4.9 N x + 9.8J - 25J

0 = 5.0 N/m x^2 + 4.9 N x - 15.2J

or in terms of J and meters

(7) 0 = 5.0 J/m^2 x^2 + 4.9 J/m x - 15.2J

or dividing through by Joules

(8)  0 = 5.0 (x/m)^2 + 4.9(x/m) -15.2

let y =x/m to give

(9)  0 = 5.0 y^2 + 4.9y -15.2

Applying the quadratic formula to this gives
______________________
(10)         -4.9 +/- \/ 4.9^2 - 4(5.0)(-15.2)
y =    ----------------------------------
2(4.9)

-4.9 +/- 18.11
=      --------------
9.8

=    (-23.01/9.8 , 13.21/9.8)

=    ( -2.34, 1.35)

x was taken to be positive so we have a comperssion of

(11)     x = 1.35m

the initial potential energy of the compressed spring is then

1
(12) PE = - k x^2 = - 10N/m (1.35m)^2 = 9.1125 J ~ 9.11J
2

When the block reverses directio it slides back toward where it came
from comming to a halt before its initial location because of
friction.  It covers the first 1.35m while the spring is still pushing
and then a length L given by

(13)   9.11J = Uk mg (L +1.35m)

= .25 2kg 9.8m/s^2 (L +1.35m)

= 4.9N L + 6.615 N m

= 4.9N L + 6.615 J

Thus

(14)     2.495 J = 4.9N L

L = 2.495 J/ 4.9N = .509 m

**********************************************
*                                            *
* spring compression = 1.35m                 *
*                                            *
* distance from spring when at rest = .509m  *
*                                            *
**********************************************
```

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