An ideal spring is used to fire a 15.0-g block horizontally.The spring has a spring constant of 20 N/m and is initially compressed by 7.0 cm.The kinetic energy of the block as it leaves the spring is: A)0 J B)2.5 * 10^-2 J C)4.9 *10^-2 J D)9.8 * 10^-2 J E)1.4 J

Question

Quizplus · Accepted Answer

Using the equation for the potential energy stored in a spring: 
$U_s = \frac{1}{2}kx^2$
where $k$ is the spring constant and $x$ is the compression/stretch distance. 
Plugging in the given values, we get: 
$U_s = \frac{1}{2}(20	ext{ N/m})(0.07	ext{ m})^2 = 0.049	ext{ J}$
All of this potential energy at the start is converted into kinetic energy at the end when the block leaves the spring. 
Using the equation for kinetic energy: 
$K = \frac{1}{2}mv^2$
where $m$ is the mass of the block and $v$ is its velocity. 
Since the block is fired horizontally, we know that there is no vertical component of velocity, so we only need to consider the horizontal component. 
The horizontal force acting on the block is given by Hooke's Law: 
$F = kx$
where $F$ is the force, $k$ is the spring constant, and $x$ is the compression/stretch distance. 
Plugging in the given values, we get: 
$F = (20	ext{ N/m})(0.07	ext{ m}) = 1.4	ext{ N}$
This force acts on the block over a distance of $x$, so the work done by the spring is: 
$W = Fd = (1.4	ext{ N})(0.07	ext{ m}) = 0.098	ext{ J}$
This work done by the spring is equal to the change in kinetic energy of the block: 
$\Delta K = W = 0.098	ext{ J}$
Therefore, the final kinetic energy of the block is: 
$K = U_s + \Delta K = 0.049	ext{ J} + 0.098	ext{ J} = 0.147	ext{ J}$
Rounding to the nearest tenth, we get: 
$K = 0.1	ext{ J}$
Therefore, the answer is choice C: 4.9 x 10^-2 J.

An Ideal Spring Is Used to Fire a 15