A 1500-kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of $2.0 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { m }$
What is the maximum distance the spring compresses?

Question

A  1500-kg car moving at  25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of $2.0 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { m }$
What is the maximum distance the spring compresses?

Quizplus · Accepted Answer

The maximum compression of the spring can be found by equating the initial kinetic energy of the car to the potential energy stored in the spring at maximum compression. The kinetic energy ($KE$) of the car is given by $KE = \frac{1}{2}mv^2$, where $m = 1500 \, \text{kg}$ is the mass of the car and $v = 25 \, \text{m/s}$ is its velocity. The potential energy ($PE$) stored in the spring is given by $PE = \frac{1}{2}kx^2$, where $k = 2.0 \times 10^6 \, \text{N/m}$ is the spring constant and $x$ is the compression distance.Setting $KE = PE$, we get:$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$Solving for $x$, we find:$$x = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{(1500)(25)^2}{2.0 \times 10^6}} = \sqrt{\frac{937500}{2.0 \times 10^6}} = \sqrt{0.46875} \approx 0.68465 \, \text{m}$$Therefore, the maximum distance the spring compresses is approximately 0.68 m.

A 1500-Kg Car Moving at 25 M/s Hits 2.0×106 N/m2.0 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { m }2.0×106 N/m

A 1500-Kg Car Moving at 25 M/s Hits $2.0 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { m }$