A bullet mass is 60g is fired at a speed of 300 m/s what speed will the bullet emerge from the other side of a 2.5 cm plate if the plate resists with a force of 3300, how much kinetic energy of the bullet decresaes over this distance?

To solve this problem, we'll calculate the bullet's final speed after it passes through the plate, and then we'll determine how much kinetic energy it loses over the distance of the plate.

### Given dataTo solve this problem, we'll calculate the bullet's final speed after it passes through the plate, and then we'll determine how much kinetic energy it loses over the distance of the plate.

### Given 
- Mass of bullet: \( m = 60 \, \text{g} = 0.060 \, \text{kg} \)
- Initial speed of bullet: \( v_i = 300 \, \text{m/s} \)
- Thickness of plate: \( d = 2.5 \, \text{cm} = 0.025 \, \text{m} \)
- Force resisting the bullet: \( F = 3300 \, \text{N} \)

### 1. Calculate the Work Done by the Force (which will reduce the bullet's kinetic energy):
Work \( W \) done by the force is given by:
\[
W = F \times d
\]
Substituting the known values:
\[
W = 3300 \, \text{N} \times 0.025 \, \text{m} = 82.5 \, \text{J}
\]

### 2. Calculate the Kinetic Energy Lost by the Bullet:
The work done by the force is equal to the kinetic energy lost by the bullet. Thus, the bullet loses 82.5 J of kinetic energy as it passes through the plate.

### 3. Calculate the Bullet's Final Kinetic Energy:
The initial kinetic energy \( KE_i \) of the bullet is given by:
\[
KE_i = \frac{1}{2} m v_i^2
\]
Substituting the known values:
\[
KE_i = \frac{1}{2} \times 0.060 \, \text{kg} \times (300 \, \text{m/s})^2 = 2700 \, \text{J}
\]
After passing through the plate, the bullet's kinetic energy decreases by 82.5 J, so the final kinetic energy \( KE_f \) is:
\[
KE_f = KE_i - W = 2700 \, \text{J} - 82.5 \, \text{J} = 2617.5 \, \text{J}
\]

### 4. Calculate the Bullet's Final Speed:
The final kinetic energy is related to the final speed \( v_f \) by:
\[
KE_f = \frac{1}{2} m v_f^2
\]
Solving for \( v_f \):
\[
v_f = \sqrt{\frac{2 KE_f}{m}} = \sqrt{\frac{2 \times 2617.5 \, \text{J}}{0.060 \, \text{kg}}}
\]
Now, let's compute the final speed.The bullet's final speed after passing through the plate is approximately \( 295.38 \, \text{m/s} \).

### Summary:
1. The bullet loses 82.5 J of kinetic energy while passing through the plate.
2. The bullet's final speed after emerging from the plate is \( 295.38 \, \text{m/s} \).

Would you like further details or explanations?

Here are 5 related questions to expand on this topic:
1. How does increasing the plate's resistance affect the bullet's final speed?
2. How would the bullet's final speed change if the plate were thicker?
3. What happens to the kinetic energy if the bullet's initial speed is doubled?
4. How does the mass of the bullet affect the amount of energy it loses?
5. Can we calculate the time the bullet takes to pass through the plate?

**Tip:** Work and energy are closely related; the work done by a force over a distance directly affects an object's kinetic energy.

A bullet mass is 60g is fired at a speed of 300 m/s what speed will the bullet emerge from the other side of a 2.5 cm plate if the plate resists with a force of 3300, how much kinetic energy of the bullet decreases over this distance?

Solve a physics problem involving a 60g bullet fired at 300 m/s, passing through a 2.5 cm thick plate with a resisting force of 3300 N. Learn how to calculate the bullet's final speed and how much kinetic energy decreases using the work-energy theorem. Suitable for Grades 10-12.

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