- #1

Nayef

- 3

- 0

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**W_{s}(work done by spring force) = -_{app}(work done by applied force) , F_{s}(spring force) = -kx , Ws = -1/2 kx^{2}**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 F**

==> F

But the answer and method according to the solution manual is different, it is given below

This is a situation where Eq W

_{app}= -F_{s}==> F

_{app}= -(-kx) ==> 3= 50x ==> x = 3/50 = 0.06But 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 haveFx = 1/2 kx

^{2}

==> 3 x = 1/2 (50 )x

^{2}

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 – 1/2kx

^{2}. 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.