Question

one end of an ideal spring is attached to a wall and the other end to a block as shown. the block is initially at equilibrium at ( x = 0 ) m. a physics student determines that stretching the spring by moving the block from ( x = 0 ) m to ( x = 0.05 ) m stores 4.0 j of energy in the block - spring system. how much additional energy is stored in the block - spring system after the block is moved from ( x = 0.05 ) m to ( x = 0.10 ) m?
a 4.0 j
b 8.0 j
c 12 j
d 16 j

Explanation:

Step1: Recall the formula for elastic potential energy

The elastic potential energy formula is \(U = \frac{1}{2}kx^{2}\), where \(U\) is the potential energy, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position.
When \(x_1=0.05\ m\), \(U_1 = 4.0\ J\). So, \(4.0=\frac{1}{2}k(0.05)^{2}\).

Step2: Find the spring constant \(k\)

From \(4.0=\frac{1}{2}k(0.05)^{2}\), we can solve for \(k\).

$$
LATEXBLOCK0
$$

Step3: Calculate the energy at \(x = 0.10\ m\)

Using \(U=\frac{1}{2}kx^{2}\) with \(k = 3200\ N/m\) and \(x = 0.10\ m\), we get \(U_2=\frac{1}{2}\times3200\times(0.10)^{2}\).

$$
LATEXBLOCK1
$$

Step4: Calculate the additional energy

The additional energy \(\Delta U=U_2 - U_1\).
Since \(U_1 = 4\ J\) and \(U_2=16\ J\), \(\Delta U=16 - 4=12\ J\)

Answer:

C. \(12\ J\)