A block lies on a horizontal frictionless surface, and the spring constant is 50 N/m. Initially, the spring is at its relaxed length and the block is station ary at position x = 0 .Then an applied force with a constant magnitude of 3 N pulls the block in the positive direction of the x axis,stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) block's position when its kinetic energy is maximum Ws(work done by spring force) = -Wapp(work done by applied force) , Fs(spring force) = -kx , Ws = -1/2 kx2 I tried to solve the solution by using the concept , that because our applied force is of constant magnitude and because the block stops after some displacement x , then at that point Fapp = -Fs ==> Fapp = -(-kx) ==> 3= 50x ==> x = 3/50 = 0.06 But the answer and method according to the solution manual is different, it is given below This is a situation where Eq Ws = -Wapp , applies, so we have Fx = 1/2 kx2 ==> 3 x = 1/2 (50 )x2 which (other than the zero) gives x = (3.0/25) m = 0.12 m. What is wrong with my logic, the second part of the question is clear to me but i am providing its answer according to the solution manual because it seems relevant part first and might help in understanding part first better. With Kf = K considered variable and Ki = 0, Eq. 7-27 gives K = Fx – 1/2kx2. We take the derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position xc which corresponds to a maximum value of K: xc = F/k= (3.0/50) m = 0.060 m.