we must apply a force of magnitude 82.0 N to hold the block stationary at x=-2.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the block is then again stationary. What are the block's positions, in cm? ((a) positive and (b) negative)
You can't know this unless you know the spring constant.
F=ma
To determine the block's positions, we can start by analyzing the work done on the spring-block system. The work done is given by the equation W = ½kx^2, where W is the work done, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the work done is +8.0 J. We can rearrange the equation to solve for x:
8.0 J = ½kx^2
Given that the force to hold the block stationary is 82.0 N, we can relate it to the spring constant using Hooke's Law:
Force = -kx
Substituting the given force and solving for k:
82.0 N = k(0.02 m)
k = 82.0 N / 0.02 m
k = 4100 N/m
Now, we can substitute the obtained values of k and the work done into the equation:
8.0 J = ½(4100 N/m)x^2
Rearranging the equation:
16 J/Nm = x^2
Taking the square root of both sides:
±√16 J/Nm = ±x
Simplifying the equation:
±4 J/N = ±x
So, the block's positions are:
(a) +4 cm
(b) -4 cm