Question

10 12 X (m)
For the arrangement of forces in the figure and the table below, a 2.00 kg particle is released at x = 5.00 m with an initial velocity of 3.42 m/s in the negative direction of the x axis. The potential energy at x = 5.00 m is zero (a) If the particle can reach x = 0 m, what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 5.00 m at speed 3.42 m/s. (b) If the particle can reach x = 13.0 m, what is its speed there, and if it cannot, what is its turning point?

Answer 1

deceleration of the particle during F2 = 5/2

=2.5 m/s^2

deceleration when F1 is applied = 3/2

=1.5 m/s^2

deceleration when F3 is applied = -4/2

=-2 m/s^2

deceleration when F4 is applied = -1/2

=-0.5 m/s^2

a)speed after crossing F2 be v.

so, v^2=u^2+2aS

or v^2 = 3.42^2 - 2*2.5*1

or v=-2.587 m/s

now during the third region, assuming that it can cross the region,

let the speed be v.so,

v^2 = u^2 + 2aS

or v^2 = 2.587^2 + 2*1.5*(-2)

or v=0.832 m/s

for the other case,

let the speed after crossing the first region be v.so,

v^2=u^2+2*a*S

or v^2 = 3.42^2 - 2*2*2

or v=1.9226 m/s

now assuming that it can cross the region,

let the speed be v.so,

v^2=u^2-2*a*s

or v^2=1.9226^2 -2*0.5*1

or v=1.642 m/s

b)the speed at x=13m is 1.642 m/s