Physics problem need help

Question

A 9.10 kg particle is subject to a net force that varies with position as shown in the figure. The particle starts moving at x = 0, very nearly from rest.

 

http://www.webassign.net/pse/p7-17.gif

 

 

What is its speed at the following positions?
(a) x = 5.00 m

 

 

(b) x = 10.0 m

 

(c) x = 15.0 m

Answer 1

This solution is long and is based on integration .acceleration = Fore / mass = 3/9.1 = 0.33 m/s2 approx

F = kX 
K = F/X = 3/X = 3/5
F = (3/5) X
a = (3xm/5) X = dV/dt
dV =(3xm/5) X dt
but dX/dt = V
so dt= dX/v
dV = (3xm/5) X dX /V
0r
V dV = 3xm/5 dX
Integral V dV = 3xm/5 dX
V2 /2 = (3m/5) X + C
V2 = 2x3 xmxX/5 +C
Assumi9ng at t= 0 V= 0
C= 0
V 2 = 6xmX/5
At X = 5 , V2 = 6m V = sqr (6x 9.1) = velocity at X= 5 ,
V final = Velocity at 5
Upto 10 m force is constant = 3N, acceleration = 3/9.1

 

 

If some one is interested I have the remaining portion also