A Mass on a Spring Problem

I have been trying to understand this Q for 2 days now, but just cant get it . Here it is Q)A block a lies on a horrizontal frictionless surface and the spring constant is 50N/m. Initially the spring is at its relaxed length and the block is stationary at position x=0. Then an applied force with a constant magnitude of 3N 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.The solution according to the manual is : Since block is stationary before & after displacement therefore W(applied force) = -W(spring force) Fx= -(-1/2 kx^2) => 3x=1/2 50x^2 ,by solving the eq root other than 0 comes out x=0.12 .I tried to solve this Q by using the concept that because our applied force is constant= 3N ,and because the block stops after some displacement x, then at that point F(applied)=F(spring) => 3=kx => 3=50x => x=0.06 .What is wrong with my logic, why do i obtain a different answer.The second part is clear to me but i have given its solution because it is relevant to the first part & might be of help in helping me understand part first.Here is the answer of second part acc to manual : With Kf = K considered variable and Ki = 0 => K = FX-1/2 kx^2(work-energy principle). We take derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position Xm which corresponds to a maximum value of K. Xm = F/k= 3/50= 0.060 m.We note that Xm is also the point where the applied and spring forces “balance.�
- Mohammad Nayef (age 18)
India

Hello Mohammad,

You state that when the spring is at its maximum extension the forces of the extended spring and the applied force are equal.   This is incorrect, the spring force is in fact double the applied force.
This problem is identical to the one where a mass in a gravitational field is suspended by a spring.   The system exhibits simple harmonic motion with amplitude 2Mg/k and ω² = k/M.

LeeH

(published on 06/09/2015)