I would like to thoroughly understand each steo to solve this.

In 0.750s, a 7.0kg block is pulled through a distance of 4.0m on a frictionless horizontal surface from rest. the block has a constant acceleration and is pulled by means of a horizontal spring attached to the block. The spring constant is 415N/m. By how much does the spring stretch?

solve for acceleration

d=1/2 a t^2 solve for a, then
f=ma solve for f.
Then, knowing F
f=kx

to solve for a

To find out how much the spring stretches, we can use the formula for the force exerted by a spring:

F = kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement (stretch) of the spring.

In this problem, we can assume that the force exerted by the spring is equal to the force required to accelerate the block. We can use Newton's second law of motion:

F = m * a

where F is the force, m is the mass of the block, and a is the acceleration.

First, let's find the acceleration of the block using the time, displacement, and mass given in the problem.

Given:
Time (t) = 0.750s
Displacement (d) = 4.0m
Mass (m) = 7.0kg

Acceleration (a) can be found using the equation:

a = d / t^2

Substituting the given values:

a = 4.0m / (0.750s)^2

Next, we can calculate the force exerted by the spring using Newton's second law:

F = m * a

Substituting the known values:

F = 7.0kg * a

Now, we can use the formula for the force exerted by the spring:

F = k * x

Rearranging the equation, we get:

x = F / k

Substituting the known values:

x = (7.0kg * a) / (415N/m)

Now we can calculate the stretch of the spring (x) by substituting the known values for mass and acceleration into the equation.

x = (7.0kg * a) / (415N/m)

Simply plug in the values of mass and acceleration that we calculated earlier to find the stretch of the spring.