Energy and Work of a bullet

In summary, the conversation discusses finding the magnitude of the force required to stop a 5.00g bullet moving at 600.0 m/s, which penetrates a tree trunk to a depth of 4.00 cm. The solution involves using the kinetic energy-work theorem and Newton's law to find the force and acceleration, respectively. The final answer obtained is 22500N.
  • #1
lilwigz
22
0

Homework Statement


A 5.00g bullet moving at 600.0 m/s penetrates a tree trunk to a depth of 4.00 cm.

A) Use work and energy considerations to find the magnitude of the force that stops the bullet.

B) Assuming that the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment the bullet stops moving.


Homework Equations


Ek= 1/2mv^2
W=fdcos(0)


The Attempt at a Solution



I tried to do Ek=1/2mv^2
Ek= 1/2 x ( .005kg) x (600m/s)^2 but i think that's wrong
 
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  • #2
Ek= 1/2 x ( .005kg) x (600m/s)^2 but i think that's wrong

This is how much energy the bullet intially has.

All this energy is transferred as the bullet travels through the tree a distance of 4cm.

So how can we solve for the magnitude of the force which stops the bullet?

HINT: Use kinetic energy-work theorem.
 
Last edited:
  • #3
with the formula Fd(cos0)= (1/2 mvf^2)-(1/2mvi^2) i got -900J/.04m and i got the answer 22500N horizontally
 
  • #4
looks good to me
 
  • #5
i got the answer 22500N horizontally

They are asking for the magnitude of the force so you don't have to indicate a direction ("horizontally" in your case).

22500N is good enough! Now try part B!
 
  • #6
since the frictional force is constant, the acceleration would be constant, so

v²=u²+2as
0=600²+2a(0.04)
a=-360000/0.08 = -4.5x10^6

a=(v-u)/t
t=(v-u)/a
=0-600/(-4.5x10^6)
=0.133 ms
 
  • #7
v²=u²+2as
0=600²+2a(0.04)
a=-360000/0.08 = -4.5x10^6

a=(v-u)/t
t=(v-u)/a
=0-600/(-4.5x10^6)
=0.133 ms

Looks good to me!
 
  • #8
You can find the acceleration a by Newton's law: F=ma

Iono how you're trying to find it there...answer looks right though.
 
  • #9
thank you!
 
  • #10
You can find the acceleration a by Newton's law: F=ma

How would you do that? You have no way of computing the frictional force.
 
  • #11
You just did in part A
...
 
  • #12
Matterwave said:
You just did in part A
...

Doh! Brainfart :)
 

1. What is the relationship between energy and work in a bullet?

The energy of a bullet is directly related to the work it can do. In simple terms, the more energy the bullet has, the more work it can do when it hits its target.

2. How is the energy of a bullet calculated?

The energy of a bullet can be calculated using the formula E = 1/2 * m * v^2, where E is energy, m is mass, and v is velocity. This formula is known as the kinetic energy formula.

3. What factors affect the energy and work of a bullet?

The energy and work of a bullet are affected by several factors, including the mass and velocity of the bullet, the type of gun used, and the distance the bullet travels.

4. Can the energy and work of a bullet be increased?

Yes, the energy and work of a bullet can be increased by increasing its velocity or mass. This can be achieved by using a more powerful gun or by using a heavier bullet.

5. How does the energy and work of a bullet affect its effectiveness?

The energy and work of a bullet directly impact its effectiveness. A higher energy and work means the bullet can penetrate deeper and cause more damage to its target, making it more lethal.

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