A 2.0-kg block sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 200 N/m) which has its other end fixed. If the block has a speed of 4.0 m/s as it passes through the equilibrium position, what is its speed when it is 20 cm from the equilibrium position?
A) 2.6 m/s
B) 3.1 m/s
C) 3.5 m/s
D) 1.9 m/s
E) 2.3 m/s

Question

Quizplus · Accepted Answer

The potential energy stored in the spring is given by:

$U=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement from the equilibrium position.

At the equilibrium position, the spring is unstretched, so the potential energy is zero:

$U=0$

When the block is 20 cm from the equilibrium position, its displacement is:

$x=0.2	ext{ m}$

The total mechanical energy of the system (block + spring) is conserved, so we can use the conservation of energy principle to find the speed of the block at this position. At the equilibrium position, the total energy is entirely kinetic:

$E_k=\frac{1}{2}mv_{eq}^2$

where $v_{eq}$ is the speed of the block at the equilibrium position.

When the block is 20 cm from the equilibrium position, the total energy is the sum of kinetic and potential energies:

$E_k + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Setting the initial and final energies equal, we have:

$\frac{1}{2}mv_{eq}^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Solving for v, we have:

$v = \sqrt{\frac{k}{m}(x_{eq}^2 - x^2) + v_{eq}^2}$

Substituting the given values, we get:

$v = \sqrt{\frac{200}{2}(0.2^2 - 0^2) + 4^2} \approx 3.5 	ext{ m/s}$

Therefore, the best choice is C.

A 20-Kg Block Sliding on a Horizontal Frictionless Surface Is Attached