A spring-powered dart gun is unstretched and has a spring constant 100 N/m. The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun. The maximum height of the projectile when it is shot straight up from the gun is
A) 7.57 m.
B) 7.00 m.
C) 6.53 m.
D) 6.00 m.
E) 5.50 m.

Question

Quizplus · Accepted Answer

The potential energy stored in the spring when it is compressed is given by:
$U_s = \frac{1}{2}kx^2$
where k is the spring constant and x is the compression of the spring.
Substituting the given values, we have:
$U_s = \frac{1}{2}(100 N/m)(0.08 m)^2 = 0.32 J$
This potential energy is converted to kinetic energy of the projectile when it is released. The kinetic energy of the projectile is given by:
$K = \frac{1}{2}mv^2$
where m is the mass of the projectile and v is its velocity.
We can use conservation of energy to find the velocity of the projectile at the instant it leaves the gun. At the maximum height, the kinetic energy of the projectile is zero and all the initial potential energy is converted to gravitational potential energy:
$U_g = mgh$
where h is the maximum height reached by the projectile.
We can equate the initial potential energy to the final gravitational potential energy:
$U_s = U_g$
$\frac{1}{2}kx^2 = mgh$
Solving for h, we get:
$h = \frac{1}{2}(\frac{kx^2}{mg})$
Substituting the given values, we have:
$h = \frac{1}{2}(\frac{(100 N/m)(0.08 m)^2}{(0.005 kg)(9.81 m/s^2)}) = 6.53 m$
Therefore, the maximum height of the projectile when it is shot straight up from the gun is 6.53 m. The answer is choice (C).

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