A 1.81-kg block slides on a horizontal frictionless table with a speed of 1.22 m/s. It is brought to rest by compressing a spring in its path. If the spring has a force constant of 7.30 N/m, it is compressed
A) 6.89 m.
B) 1.22 m.
C) 1.71 m.
D) 0.369 m.
E) 0.607 m.

Question

Quizplus · Accepted Answer

The initial kinetic energy of the block is equal to the potential energy stored in the compressed spring. Therefore, we can use the equation for potential energy of a spring:
$U =\dfrac{1}{2}kx^2$
where $k$ is the spring constant and $x$ is the compression distance.
The initial kinetic energy of the block is:
$K = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(1.81 	ext{ kg})(1.22 	ext{ m/s})^2 = 1.394 	ext{ J}$
This is equal to the potential energy stored in the compressed spring:
$K = U = \dfrac{1}{2}kx^2$
Plugging in the known values, we get:
$1.394 	ext{ J} = \dfrac{1}{2}(7.30 	ext{ N/m})(x)^2$
Solving for $x$, we get:
$x = \sqrt{\dfrac{2(1.394 	ext{ J})}{7.30 	ext{ N/m}}} = 0.607 	ext{ m}$
Therefore, the block compresses the spring by 0.607 m before coming to a stop. The answer is (E).

A 181-Kg Block Slides on a Horizontal Frictionless Table with a a Speed