A 0.400 kg block of wood hangs from the ceiling by a string, and a 0.0700 kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.80 m/s. The wad of putty sticks to the block.
Is the mechanical energy of this system conserved?
yes or no
(b) How high does the putty-block system rise above the original position of the block?
____ cm
Energy is not conserved.
momentum is conserved.
momentum putty=momentumclump
.07*5.8=(.4+.07)Vi solve for Vi upward.
Vi^2=(.4+.07)g h solve for h.
What’s up g. I’m not sure where you got this from (Vi^2=(.4+.07)g h solve for h.).
Mgh is used to compute potential energy, so I don’t understand why you set it equal to Vi^2. Oh btw I noticed that you posted this in 2010 so you probably won’t even see this.
To determine if the mechanical energy of the system is conserved, we need to consider whether there are any external forces acting on the system that could do work. In this case, the only external force acting on the system is gravity, which does work as the block and putty rise.
(a) Is the mechanical energy of this system conserved?
No, the mechanical energy of the system is not conserved because there is an external force (gravity) doing work on the block and putty.
To calculate the height the putty-block system rises above the original position of the block, we can use the principle of conservation of mechanical energy, assuming no energy is lost to other forms like sound or heat. We can equate the initial mechanical energy to the final mechanical energy:
Initial mechanical energy = Final mechanical energy
The initial mechanical energy can be calculated as the potential energy of the block (mgh), where m is the mass of the block, g is the acceleration due to gravity, and h is the height above the original position of the block.
The final mechanical energy can be calculated as the potential energy of the block and putty system, given that they rise together. The mass of the system is the sum of the masses of the block and putty.
Final mechanical energy = potential energy of block and putty
= (m_block + m_putty) * g * h_final
Since the putty sticks to the block, the final velocity of the combined system is zero. Using the principle of conservation of momentum, we can calculate the final height (h_final) by equating the initial momentum to the final momentum:
Initial momentum = Final momentum
m_putty * v_putty_initial = (m_block + m_putty) * 0
Simplifying the equation, we get:
v_putty_initial = 0 = (m_block + m_putty) * v_final
Solving for v_final, we find:
v_final = v_putty_initial / (m_block + m_putty)
Now, we can calculate the final mechanical energy:
Final mechanical energy = (m_block + m_putty) * g * h_final
Setting the initial and final mechanical energies equal to each other, we have:
m_block * g * h = (m_block + m_putty) * g * h_final
Simplifying the equation and solving for h_final, we get:
h_final = (m_block * h) / (m_block + m_putty)
Now we can substitute the given values:
m_block = 0.400 kg
m_putty = 0.0700 kg
h = unknown
h_final = unknown
Substituting these values into the equation, we can solve for h_final, which will give us the height the putty-block system rises above the original position of the block.
To determine if the mechanical energy of the system is conserved, we need to consider the initial and final states of the system.
Initially, the block of wood is at rest, hanging from the ceiling. The wad of putty is thrown straight upward and strikes the bottom of the block, sticking to it.
In the final state, the block and putty together rise above their original position.
Since there are external forces acting on the system (gravity and tension in the string), mechanical energy is not conserved.
To find out how high the putty-block system rises above the original position of the block, we can use the principle of conservation of momentum.
The momentum before the collision is given by:
m1 * v1 = m1 * u1 + m2 * u2,
where
m1 = mass of the block = 0.400 kg,
v1 = velocity of the block just before impact = 0 m/s (since it's at rest),
m2 = mass of the putty = 0.0700 kg,
u1 = velocity of the block after impact (which we need to find),
u2 = velocity of the putty before impact = 5.80 m/s.
Since the putty sticks to the block, the final velocity of the block and putty system (u1) would be the same. We can rewrite the equation as:
0.400 kg * 0 m/s = (0.400 kg + 0.0700 kg) * u1.
Simplifying the equation gives:
0 = 0.4700 kg * u1.
Solving for u1 gives:
u1 = 0 m/s.
This means that after the collision, the putty-block system comes to a stop momentarily before starting to rise.
To determine how high the system rises, we can now use the principle of conservation of energy:
mgh = (1/2)mv^2,
where
m = total mass of the system = 0.400 kg + 0.0700 kg = 0.4700 kg,
g = acceleration due to gravity = 9.8 m/s^2,
h = height the system rises above the original position,
v = final velocity of the system = 0 m/s.
Since the final velocity is zero, the equation becomes:
mgh = 0.
This means that the system does not rise above its original position. Hence, the height (h) is 0 cm.
In summary:
(a) The mechanical energy of the system is not conserved. (Answer: No)
(b) The putty-block system does not rise above the original position of the block. (Answer: 0 cm)