# Homework Help: Spring on inclined plane with a Block

1. Apr 5, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
A block of mass m = 4 kg, is held at rest on a spring (point A), spring constant k =
500 N / m, compressed by a distance AB = .DELTA.L as shown in Figure 3. When the block is freed ,
it reaches the point B ( non-deformed position of the spring ) with a velocity of VB = 5 m / s. assume
the coefficient of kinetic friction between the block and the incline is μk = 0.15 , determine :
a) compression of the spring ;
b) the distance traveled by the block to stop, point C (measured from point B);

Problem original drawing:
http://i.imgur.com/i98xq8J.png

2. Relevant equations
∑Fy = m*ay
Fk = μk * N
Wi->f = F(cos(angle)*DELTA X)
Ugi + Uei + (Wi->f) + 1/2 *m*vi2 = Ugf + Uef + 1/2*m*vf2

3. The attempt at a solution
My drawing of the problem:
The spring become uncompressed at B.
http://i.imgur.com/Db1HwA0.jpg

v = speed (m/s)

∑Fy = m*ay
ay = 0 // m = 4kg
N - W*cos(35) = m*ay => N - (4*9.81)*cos(35) = 4*0
N = 32.1435N

Fk = μk * N
Fk = 0.15 * 32.1435N
Fk = 4.82153N

Wi->f = F(cos(angle)*DELTA X)
Wi->f = 4.82153N(cos(35)*DELTA X)

Ugi + Uei + (Wi->f) + 1/2 *m*vi2 = Ugf + Uef + 1/2*m*vf2

Ugi = 0J
Uei = 1/2 * K * Xsf2
Wi->f = 4.82153N(cos(35)*DELTA X)
1/2 *m*vi2 = 0 since speed initial is 0m/s
Ugf = m*g*hf = 4Kg*9.81* hf*sin(35)
Uef = 0J
1/2*m*vf2 = 1/2*4Kg*5m/s2

I am stuck here.... :(

Last edited: Apr 5, 2015
2. Apr 5, 2015

### SammyS

Staff Emeritus
You can post those links as images if they're not too large, like I did with one of yours