A 4.0 kg block is initially at rest on a frictionless, horizontal surface. The block is moved 8.0 m by the application of a constant 10.0 N horizontal force. If the block slides into a fixed horizontal spring and comes to rest when the spring is compressed a distance of x=0.25m. Determine the spring constant of the fixed horizontal spring. Show all formula with substitutions and units.

To find the spring constant, you take into account that all the kinetic energy of the block becomes elastic potential energy in the spring, when the block compressed totally the spring:

[tex]K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]

m: mass of the block = 4.0kg

v: velocity of the block just before it hits the spring

x: compression of the spring = 0.25m

k: spring constant = ?

You solve the previous equation for k:

[tex]k=\frac{mv^2}{x^2}[/tex] (1)

Then, you have to calculate the velocity v of the block. First, you calculate the acceleration of the block by using the second Newton law:

[tex]F=ma[/tex]

F: force over the block = 10.0N

a: acceleration

[tex]a=\frac{F}{m}=\frac{10.0N}{4.0kg}=2.5\frac{m}{s^2}[/tex]

With this value of a you can calculate the final velocity after teh block has traveled a distance of 8.0m:

[tex]v^2=v_o^2+2ad[/tex]

vo: initial velocity = 0m/s

d: distance = 8.0m

[tex]v=\sqrt{2ad}=\sqrt{2(2.5m/s^2)(8.0m)}=6.32\frac{m}{s}[/tex]

Now, you can calculate the spring constant by using the equation (1):

[tex]k=\frac{mv^2}{x^2}=\frac{(4.0kg)(6.32m/s)^2}{(0.25m)^2}=2560\frac{N}{m}[/tex]

hence, the spring constant is 2560 N/m