A spring-powered dart gun is unstretched and has a spring constant 12.0 N/m. The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun. The velocity of the projectile when it is shot from the gun is
A) 1.52 m/s.
B) 2.54 m/s.
C) 3.92 m/s.
D) 4.24 m/s.
E) 5.02 m/s.

Question

Quizplus · Accepted Answer

The velocity of the projectile can be found using the conservation of energy principle. The potential energy stored in the compressed spring is converted into the kinetic energy of the projectile. The potential energy ($PE$) stored in the spring when compressed is given by the formula $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the compression distance. The kinetic energy ($KE$) of the projectile when it is shot is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the projectile and $v$ is its velocity. Setting the potential energy equal to the kinetic energy and solving for $v$ gives:$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$Given $k = 12.0\, N/m$, $x = 8.00\, cm = 0.08\, m$, and $m = 5.00\, g = 0.005\, kg$, we can solve for $v$:$$12.0 \times 0.08^2 = 0.005 \times v^2$$$$v^2 = \frac{12.0 \times 0.08^2}{0.005}$$$$v = \sqrt{\frac{12.0 \times 0.08^2}{0.005}} \approx 3.92\, m/s$$Therefore, the velocity of the projectile is approximately 3.92 m/s.

A Spring-Powered Dart Gun Is Unstretched and Has a Spring