Question

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.

Answer 1

Answer:

k = 2560 N/m

Explanation:

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:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$

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:

$k=\frac{mv^2}{x^2}$     (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:

$F=ma$

F: force over the block = 10.0N

a: acceleration

$a=\frac{F}{m}=\frac{10.0N}{4.0kg}=2.5\frac{m}{s^2}$

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

$v^2=v_o^2+2ad$

vo: initial velocity = 0m/s

d: distance = 8.0m

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

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

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

hence, the spring constant is 2560 N/m