A bullet with the mass of 45 g is fired into an 8.3 kg block of wood resting on a floor against a spring. This ideal spring (k = 76 N/m) has a maximum compression of 28 cm. What was the initial speed of the bullet?

what is the spring energy when compressed?

energy= 1/2 k * .28^2
what what the KE of the block and bullet? Same. What was then the velocity of the combination?
1/2 masstotal*v^2=1/2 k .28^2 solve for v.
But the law of conservation of momentum applied when the bullet hit the block:
massbullet*Vb=masstotal*Vabove
solve for Vb

THANK YOU. ;A;

I undergravd nothing IM 13

To find the initial speed of the bullet, we can use the principles of conservation of momentum and conservation of energy.

First, let's define our variables:
m1 = mass of the bullet (45 g = 0.045 kg)
m2 = mass of the wood block (8.3 kg)
k = spring constant (76 N/m)
x = maximum compression of the spring (28 cm = 0.28 m)
v1 = initial velocity of the bullet (what we want to find)
v2 = final velocity of the combined bullet and wood block

Conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision:
m1 * v1 = (m1 + m2) * v2

Conservation of energy tells us that the initial kinetic energy of the bullet is equal to the elastic potential energy stored in the compressed spring:
(1/2) * m1 * v1^2 = (1/2) * k * x^2

Now, let's solve the equations:

From conservation of momentum:
m1 * v1 = (m1 + m2) * v2
v2 = (m1 * v1) / (m1 + m2)

From conservation of energy:
(1/2) * m1 * v1^2 = (1/2) * k * x^2
v1^2 = (k * x^2) / m1

Substituting the value of v2 in terms of v1:
v2 = (m1 * v1) / (m1 + m2)

Substituting the values of m1, m2, k, and x:
v1^2 = (76 N/m * 0.28 m^2) / 0.045 kg

Simplifying:
v1^2 = 475.11 m^2/s^2

Taking the square root of both sides:
v1 = √475.11 m/s

Therefore, the initial speed of the bullet is approximately 21.8 m/s.