Work-Energy principle and spring force

[Nayef]

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

W_s(work done by spring force) = -W_app(work done by applied force) , F_s(spring force) = -kx , Ws = -1/2 kx²I 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 F_app = -F_s
==> F_app = -(-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 W_s = -W_app , 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:
x_c = F/k= (3.0/50) m = 0.060 m.

 

[ehild]

Welcome to PF!

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

W_s(work done by spring force) = -W_app(work done by applied force) , F_s(spring force) = -kx , Ws = -1/2 kx²I 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 F_app = -F_s
==> F_app = -(-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 W_s = -W_app , applies, so we have
==> 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:
x_c = F/k= (3.0/50) m = 0.060 m.

The block stops (momentarily) means that its velocity is zero at that moment. But it does not mean that the net force is zero; the acceleration is zero instead.
When only a constant force acts on the block, connected to the spring, it will perform SHM when no dissipative force is present. When it has maximum kinetic energy, its acceleration is zero, and when the kinetic energy is zero, the magnitude of acceracion is maximum.
You can do a little experiment: gravity also means a constant force. Hang some weight on an elastic spring. Holding the other end of the spring, let the weight fall : the weight will vibrate between a maximum and minimum height.