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)0.305 m
E)by none of these amounts

Question

Quizplus · Accepted Answer

The compression of the spring can be found using the conservation of energy principle. The initial kinetic energy of the block is converted into the potential energy stored in the spring. The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the block and $v$ is its velocity. The potential energy (PE) stored in the spring is given by $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the compression of the spring. Setting these equal to each other, we get $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$. Solving for $x$ gives $x = \sqrt{\frac{mv^2}{k}}$. Plugging in the given values ($m = 1.81 \, \text{kg}$, $v = 1.22 \, \text{m/s}$, $k = 7.30 \, \text{N/m}$), we find that none of the provided options correctly matches the calculated compression, indicating that the correct answer is not listed among the options provided.

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