Bullet strikes block on friction surface

[cashflow]

Homework Statement

A 26.2 g bullet is fired horizontally into a
1.32 kg wooden block resting on a horizon-
tal surface (μ = 0.173). The bullet goes
through the block and comes out with a speed
of 298 m/s.

If the block travels 3.01 m before coming to
rest, what was the initial speed of the bullet?
The acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s.

Homework Equations

W = μmg * d
W = .5mv_f²-.5mv_i²

The Attempt at a Solution

So I thought that the kinetic energy transferred by the bullet to the block would be the force of friction of the block.
W = .173 * (1.32 kg + .0262 kg) * 9.8 m/s² * 3.01 m = 6.869 J

6.869 J = .5(.0262 kg)(298 m/s)²-.5(.0262 kg)V_i²
= 298.87 m/s

However, my concept is likely wrong because that's not the right answer. I would appreciate any help.

 

[PhanthomJay]

Homework Statement

A 26.2 g bullet is fired horizontally into a
1.32 kg wooden block resting on a horizon-
tal surface (μ = 0.173). The bullet goes
through the block and comes out with a speed
of 298 m/s.

If the block travels 3.01 m before coming to
rest, what was the initial speed of the bullet?
The acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s.

Homework Equations

W = μmg * d
W = .5mv_f²-.5mv_i²

The Attempt at a Solution

So I thought that the kinetic energy transferred by the bullet to the block would be the force of friction of the block.
W = .173 * (1.32 kg + .0262 kg) * 9.8 m/s² * 3.01 m = 6.869 J

this is the work done by friction, and esentially correct, except eliminate the bullet mass, it is not part of the block as it starts to move.

6.869 J = .5(.0262 kg)(298 m/s)²-.5(.0262 kg)V_i²
= 298.87 m/s

However, my concept is likely wrong because that's not the right answer. I would appreciate any help.

The work done by friction is the change in KE of the BLOCK, not the bullet, you are using the wrong mass and speeds. the final speed of the block is 0, he initial speed of the block you need to solve using the other info on friction force and distance travelled. Then use conservation of momentum, where the initial speed of the block calculated avove is the speed of the block after the bullet collision.

 

[venkatg]

We do not know if the collision is elastic or inelastic. You must use the momentum conservation. I worked out and found initial velocity of the bullet to be 460m/sec. I put this in the K.E equation and found the missing energy of 1614 J that was spent by the bullet in tearing the block