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 Multiple Choice 3.92 m/s. 2.54 m/s. 1.52 m/s. 4.24 m/s. 5.02 m/s.
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![**Problem Statement:**
A spring-powered dart gun is unstretched and has a spring constant of 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:
**Multiple Choice Options:**
- ○ 3.92 m/s
- ○ 2.54 m/s
- ● 1.52 m/s
- ○ 4.24 m/s
- ○ 5.02 m/s
**Explanation:**
This problem requires understanding of the conservation of energy in mechanical systems. The potential energy stored in the compressed spring is converted into the kinetic energy of the projectile. The solution involves calculating the initial potential energy using the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k = 12.0 \, N/m \) and \( x = 0.08 \, m \).
This potential energy is then set equal to the kinetic energy of the projectile:
\[ KE = \frac{1}{2} m v^2 \]
where \( m = 0.005 \, kg \).
Solving for \( v \), the velocity of the projectile, will determine the correct option.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46e6ac57-8649-4ba4-8fc5-346c37191451%2F8350c4de-d447-4389-9a88-54c6ad623b76%2Fh79docq_processed.jpeg&w=3840&q=75)
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