Physics : Work Energy Power Question

In summary, a boy holds a 40-N weight at arm's length for 10 seconds, with his arm 1.5m above the ground. The work done by the force of the boy on the weight can be calculated by multiplying the force (40N) by the distance (1.5m) and the time (10 seconds), resulting in 600 Joules of work.
  • #1
Gunman
25
0

Homework Statement



A bullet of mass 0.0035 kg is shot into a wooden block of mass 0.121 kg.

They rise to a final height of 0.547 m as shown. What was the initial speed (in m/s) of the bullet before it hit the block?

http://www.physics247.com/physics-homework-help/conservation_nrg_quiz1.php" [Broken]

There is a picture there on how the setting looks like.


Homework Equations





The Attempt at a Solution



K.E of bullet = P.E of the total mass at its final height
1/2mv(square) = (m+M)gh
(1/2)(0.0035)v(square) = (0.0035 + 0.121)(9.81)(0.547)
v = 19.5m/s

The actual answer is 116.5m/s.

Where did I go wrong? :confused:

Ya. Thanks for any help provided. And since this is my first time posting forgive me if there are any mistakes in the way I posted the question. Thankx again. =)
 
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  • #2
Gunman said:
K.E of bullet = P.E of the total mass at its final height
Here's where you went wrong: Mechanical energy is not conserved during the collision (some of the bullet's kinetic energy becomes thermal and deformation energy). But another physical quantity is conserved during any collision--what is that?

After the collision, the mechanical energy of "bullet + block" is conserved.
 
  • #3
Thanks man. =) I got it. So the momentum of the system is conserved. So by using that I can solve this. Hm..Thank you very much for clearing my misconception. :)
 
  • #4
please help me answer this question..
A boy holds a 40-N weight at arm's length for 10 seconds. His arm is 1.5m above the ground.The work done by the force of the boy on the weight while he is holding it is?
 
  • #5


I can see that you have correctly used the conservation of energy principle in your attempt at a solution. However, there are a few things to consider in this problem.

Firstly, you have assumed that the bullet and block rise together to a final height of 0.547 m. This is not necessarily the case - the block may actually rise to a greater height than the bullet due to the transfer of momentum and energy.

Secondly, you have not taken into account the fact that the bullet will also have rotational kinetic energy as it travels through the block. This will contribute to the total kinetic energy of the system and affect the final speed of the bullet.

Finally, it is important to consider the type of collision between the bullet and the block. Is it elastic or inelastic? This will affect the amount of kinetic energy lost during the collision and therefore the final speed of the bullet.

To accurately solve this problem, you will need to take into account these factors and use the appropriate equations and principles. I would recommend consulting your textbook or a reliable online resource for a more detailed explanation and solution.
 

1. What is the formula for calculating work?

The formula for calculating work is W = F * d, where W represents work, F represents force, and d represents displacement.

2. How is power related to work and time?

Power is the rate at which work is done, and is calculated by dividing work by time. The formula for power is P = W/t, where P represents power, W represents work, and t represents time.

3. What is the difference between kinetic and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state.

4. Can energy be created or destroyed?

According to the law of conservation of energy, energy can neither be created nor destroyed, it can only be transformed from one form to another.

5. How does work-energy theorem relate work and kinetic energy?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This means that when work is done on an object, its kinetic energy will change by the same amount.

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