A block of mass 6 kg , initially at rest at the origin, is acted upon by a variable force F = 6x directed along the positive x -axis, where F is in newtons and x is in meters. The speed of the block when it reaches x = 4 m is

Step-by-step solution

The work done by the variable force is calculated by integrating the force with respect to position: W = ₀⁴ F , dx = ₀⁴ 6x , dx W = [ 3x^2 ]₀⁴ = 3(4)^2 - 0 = 48 J According to the work-energy theorem, the net work done on the block is equal to its change in kinetic energy: W = K = 1 2 mv^2 - 1 2 mu^2 Since the block starts from rest, u = 0 : 48 = 1 2 (6)v^2 - 0 48 = 3v^2 v^2 = 16 v = 4 m/s Answer: 4 m/s