In 0.780 s, a 6.50-kg block is pulled through a distance of 3.85 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 455 N/m. By how much does the spring stretch?
To find out how much the spring stretches, we can use the equation for the force exerted by a spring, which is given by Hooke's Law:
F = -kx
Where:
F is the force applied by the spring,
k is the spring constant, and
x is the displacement/stretch of the spring.
In this case, the force applied by the spring is equal to the mass of the block multiplied by its acceleration (F = ma) since there is no net external force acting on the block.
So, we have:
F = ma = -kx
We can rearrange this equation to solve for x:
x = -(ma) / k
We are given the mass of the block (m = 6.50 kg) and the spring constant (k = 455 N/m). To find the acceleration, we can use the kinematic equation:
v = u + at
Where:
v is the final velocity (which is 0 m/s since the block comes to rest),
u is the initial velocity (which is 0 m/s since the block starts from rest),
a is the acceleration of the block, and
t is the time taken (which is given as 0.780 s).
Rearranging the equation, we have:
a = (v - u) / t
= (0 - 0) / 0.780
= 0
Since the acceleration is 0, we can substitute it into the equation for x:
x = -(ma) / k
= -(6.50 kg * 0) / 455 N/m
= 0
Therefore, the spring does not stretch at all (x = 0) in this scenario.