we mustapply a force ofmagnitude80 Nto hold the block stationary at x = −2.0 cm. From that position, we then slowly move the block so that ourforce does +4.0 J of work on the spring–block system; theblock isthen again stationary.What is the block’s position?(Hint:There are two answers.)

To solve this problem, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the block-spring system is given as +4.0 J, and since the block is stationary before and after the work is done, the change in kinetic energy is zero. Therefore, the work done is equal to the change in potential energy of the spring.

The potential energy stored in a spring is given by the equation U = 1/2 kx², where k is the spring constant and x is the displacement from the equilibrium position. We can find the spring constant by using the given force and displacement when the block is stationary. The force needed to hold the block stationary is equal to the force exerted by the spring, which is given by F = kx. So, k = F/x = 80 N / 0.02 m = 4000 N/m.

Now, we can set the change in potential energy equal to the work done to find the final displacement. The initial potential energy is U_i = 1/2 kx_i² = 1/2 * 4000 N/m * (0.02 m)² = 0.8 J. The final potential energy is U_f = U_i + W = 0.8 J + 4.0 J = 4.8 J. Setting this equal to 1/2 kx_f² and solving for x_f gives x_f = sqrt((2U_f)/k) = sqrt((24.8 J)/(4000 N/m)) = 0.049 m or -0.049 m.

So, the block's final positions are x = 0.049 m and x = -0.049 m.