Question

what is the blocks speed just as it reaches the spring?

what is an expression for the change in spring potential energy, \\( \delta p e _ { s } \\) ?
a. \\( \frac { 1 } { 2 } k x _ { i } ^ { 2 } \\)
b. \\( \frac { 1 } { 2 } k x _ { f } ^ { 2 } \\)
c. \\( \frac { 1 } { 2 } k ( x _ { f } - x _ { i } ) ^ { 2 } \\)
d. \\( - \frac { 1 } { 2 } k x _ { i } ^ { 2 } \\)
e. \\( - \frac { 1 } { 2 } k x _ { f } ^ { 2 } \\)

Explanation:

Step1: Recall the formula for spring potential energy

The formula for spring potential energy is \( PE_s=\frac{1}{2}kx^{2}\), where \(k\) is the spring - constant and \(x\) is the displacement from the equilibrium position.

Step2: Calculate the change in spring potential energy

The change in spring potential energy \(\Delta PE_s = PE_{s,f}-PE_{s,i}\). Since \(x_i = 0\) (initial position of the spring is at equilibrium, so \(PE_{s,i}=\frac{1}{2}kx_{i}^{2}=0\)), and \(PE_{s,f}=\frac{1}{2}kx_{f}^{2}\). Then \(\Delta PE_s=\frac{1}{2}kx_{f}^{2}-0=\frac{1}{2}kx_{f}^{2}\)

Answer:

B. \(\frac{1}{2}kx_{f}^{2}\)