Solving Kinetic Energy of Block After Bullet Passes Through

[DPatel304]

Stuck on this one problem which states:
A bullet with a mass of 6.00 g, traveling horizontally with a speed of 370 m/s, is fired into a wooden block with mass of 0.750 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 m/s. The block slides a distance of 49.0 cm along the surface from its initial position.
Use g = 9.81 m/s^2 for the acceleration due to gravity.And I am trying to find this:
What is the kinetic energy of the block at the instant after the bullet passes through it?

If more information is needed, I can post it. Thanks in advance for your help.

Btw, the first portion of the question asked for the coefficient of friction, which I was able to get:
What is the coefficient of kinetic friction between block and surface?
0.416

and this:
What is the decrease in kinetic energy of the bullet?
368 J

 

[Hootenanny]

Can you find an expression for conservation of energy of the system?

Of course we are ignoring any deformation of the block or bullet.

 

[thomate1]

I would approach this problem by first identifying the relevant equations and principles that can be applied to solve it. In this case, we can use the conservation of momentum and the work-energy theorem.

Using the conservation of momentum, we can determine the initial velocity of the bullet before it entered the block. We know that the total momentum before and after the collision must be equal, so we can set up the following equation:

m_bullet * v_bullet = (m_bullet + m_block) * v_block

Where m_bullet is the mass of the bullet, v_bullet is its initial velocity, m_block is the mass of the block, and v_block is its velocity after the collision.

Substituting the given values, we get:

(6.00 g) * (370 m/s) = (6.00 g + 0.750 kg) * (120 m/s)

Solving for v_bullet, we get:

v_bullet = 370 m/s * (6.00 g + 0.750 kg) / 6.00 g

v_bullet = 370 m/s * (0.00600 kg + 0.750 kg) / 0.00600 kg

v_bullet = 3.75 m/s

Next, we can use the work-energy theorem to determine the decrease in kinetic energy of the bullet as it passes through the block. The work done by the force of friction (which is the only external force acting on the bullet-block system) is equal to the change in kinetic energy of the bullet. We can express this as:

W = ΔKE_bullet = ½ * m_bullet * (vf_bullet)^2 - ½ * m_bullet * (vi_bullet)^2

Where W is the work done by friction, ΔKE_bullet is the change in kinetic energy of the bullet, m_bullet is its mass, vf_bullet is its final velocity, and vi_bullet is its initial velocity.

Substituting the given values, we get:

W = ½ * (6.00 g) * (120 m/s)^2 - ½ * (6.00 g) * (370 m/s)^2

W = 368 J

Therefore, the decrease in kinetic energy of the bullet is 368 J.

Finally, to determine the kinetic energy of the block after the bullet passes through it, we can use the work-energy theorem again. This time, the work done by the force of