Calculating Kinetic Energy & Average Force of a 1.35g Bullet

In summary, the conversation is discussing the kinetic energy and average force of a bullet leaving the barrel of a gun. The kinetic energy is calculated by converting the mass of the bullet to kilograms and multiplying it by the velocity of 270m/s, resulting in an answer of 49.2075 J. The average force exerted on the bullet to move it the length of the barrel is not clear, as the individual in the conversation attempted to calculate it using mass and velocity but was unsure if it was the correct method. The concept of conservation of mechanical energy is also brought up, suggesting that the work done on the bullet should equal the change in kinetic energy if there are no losses due to friction.
  • #1
Rawpcgamer
2
0
Part 1.) 1.35g bullet leaves the barrel of the gun going 270m/s. What is its kinetic Energy?
-I converted 1.35 g to kg = 0.00135g
-I arrived with an answer of 49.2075 J

Part 2.) If the length of the barrel of the gun described above is 35cm, find the average force exerted on the bullet to move it the length of the barrel.



What I had attempted to do is Mass X Velocity/ 35 cm
-im not sure if that's the way to find avergae force, because other examples I've seen its divided by time.

thank you for looking at this for me
 
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  • #2
You know the total energy is the kinetic energy right? The work done by a force is just force*distance.
 
  • #3
im sorry, i understand what your saying but i need an thorough explanation
 
  • #4
Rawpcgamer said:
im sorry, i understand what your saying but i need an thorough explanation

If you assume no losses due to friction, then by conservation of mechanical energy, shouldn't the work done = change in kinetic energy?
 
  • #5



Hello,

I would like to provide a response to your question regarding calculating the kinetic energy and average force of a 1.35g bullet.

Part 1.) To calculate the kinetic energy of the bullet, we can use the formula KE = 1/2 * m * v^2, where m is the mass of the bullet and v is the velocity. As you correctly converted the mass to kilograms, the calculation would be KE = 1/2 * 0.00135 kg * (270 m/s)^2 = 49.2075 J. Therefore, the kinetic energy of the bullet is 49.2075 Joules.

Part 2.) To find the average force exerted on the bullet to move it the length of the barrel, we can use the formula F = m * a, where m is the mass of the bullet and a is the acceleration. In this case, the acceleration is equal to the change in velocity divided by the time taken. As we do not have the time taken, we can use the formula a = v/t, where v is the velocity and t is the time taken. The velocity of the bullet is 270 m/s and the length of the barrel is 35 cm or 0.35 m. Therefore, t = 0.35 m / 270 m/s = 0.0013 s. Now, we can calculate the acceleration: a = 270 m/s / 0.0013 s = 207692.3 m/s^2. Finally, we can calculate the average force: F = 0.00135 kg * 207692.3 m/s^2 = 280.1 N. Therefore, the average force exerted on the bullet to move it the length of the barrel is 280.1 Newtons.

I hope this helps to clarify your doubts. Please let me know if you have any further questions. Thank you for your inquiry.

Best regards,
 

1. What is the formula for calculating kinetic energy?

The formula for calculating kinetic energy is: KE = 1/2 * m * v^2, where m represents the mass of the object and v represents the velocity of the object.

2. How do you calculate the average force of a bullet?

The average force of a bullet can be calculated using the formula: F = m * a, where m represents the mass of the bullet and a represents the acceleration of the bullet. The acceleration can be calculated by dividing the change in velocity by the time it took to travel that distance.

3. What is the mass of a 1.35g bullet?

The mass of a 1.35g bullet is 0.00135 kilograms.

4. How do you convert grams to kilograms?

To convert grams to kilograms, you divide the number of grams by 1000. In this case, 1.35g would be equal to 0.00135 kilograms.

5. Does the velocity of the bullet affect the kinetic energy?

Yes, the velocity of the bullet has a direct impact on its kinetic energy. As the velocity increases, the kinetic energy also increases, and vice versa.

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