In 0.511 s, a 13.1-kg block is pulled through a distance of 3.13 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 466 N/m. By how much does the spring stretch?
F=ma=kx
k=ma/x
so what is acceleration a?
vaverage=3.13m/.511s
vfinal then must be twice that, or 6.26/.511
but we know
Vfinal^2=Vi^2+2ad so solve for a
Then, solve for k from k=ma/x
To find the amount by which the spring stretches, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = k * x
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
From the given information, we know the following:
The mass of the block (m) = 13.1 kg
The spring constant (k) = 466 N/m
The distance traveled by the block (d) = 3.13 m
We also know that the block starts from rest and has a constant acceleration. Therefore, we can use the kinematic equation to find the acceleration (a) of the block:
d = (1/2) * a * t^2
Where:
d is the distance traveled by the block,
a is the acceleration of the block, and
t is the time taken.
Rearranging the equation, we get:
a = (2 * d) / t^2
Now, we can calculate the acceleration of the block:
a = (2 * 3.13 m) / (0.511 s)^2
a ≈ 24.006 m/s^2
Now, let's calculate the force exerted by the spring using Hooke's Law:
F = k * x
Rearranging the equation to solve for x:
x = F / k
Substituting the values we have:
x = m * a / k
x = (13.1 kg) * (24.006 m/s^2) / (466 N/m)
x ≈ 0.676 m
Therefore, the spring stretches by approximately 0.676 meters.