Why is F=ma not applicable in this work and kinetic energy problem?

[adkinje]

I have been annoyed by a problem that I can't figure out. The topic is work and kinetic energy (chapter 6 in the sixth edition of physics for scientist and engineers by Paul A. Tipler). Problem 63 pg. 199

A single horizontal force F acts in the +x direction on a mass m. The intial velocity is zero, the velocity is given as a function of x v=Cx where C is a contant. If the starting point x=0 and the final point is x = x' find the force and the work.

my solution is w= .5mv^2 = .5m(Cx')^2 Therefore Fx' = .5m(Cx')^2 therefore
F=.5mx'C^2 which is the correct answer.

However, first I tried to do this

f=ma = m (dv/dt) = m[C*dx/dt] = mCx' Why is this wrong?

 

[Defennder]

x' in the book doesn't refer to dx/dt. It just refers to the final destination of the mass.

 

[baba89]

As a scientist, it is important to understand and acknowledge when our initial approach or assumptions may have been incorrect. In this case, your initial approach using F=ma is not applicable because it assumes a constant acceleration, which is not the case in this problem. Instead, the correct approach is to use the work-energy theorem, which states that the work done by a force is equal to the change in kinetic energy. In this problem, the initial velocity is zero and the final velocity is given as a function of x, so we can use the work-energy theorem to find the force and work done. It is important to always carefully consider the assumptions and principles involved in solving a problem in order to arrive at the correct solution. Keep up the good work in your studies of work and kinetic energy!