Question

A one dimensional force is acting on a particle of mass 2 kg. The force is given by F = 4 |x| where x is the x coordinate of the particle. It starts from x = -4 m with an initial velocity of 5 m/s and reaches x = + 4m with velocity of v m/s. What is v?

A) 5 m/s

(C) 4.5 m/s approx.

(B) 9.5 m/s approx.

(D) 12.4 m/s approx.

Answer 1

Target variable

v at x = +4 m

Relevant equations

v^2 = u^2 + 2ax, F = ma

Initial conditions

m = 2 kg

x0 = -4 m

F = 4x N

u = 5 m/s

Calculation

Displacement, 

x = 4 - -4 = 8 m

Acceleration, 

a = F/m = 4/2 = 2x m/s/s

Final velocity at x = 4 m,

v = sqrt(u^2 + 2ax) = sqrt(5^2 + 2*8*2x) = sqrt(25 + 32(4)) = sqrt(153) = 12.36931... m/s

Answer 

D

Comments

B is correct answer