A 9.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping.

147.77 m/s

Explanation:

I believe the question you're trying to ask is, "what is the initial speed of the bullet"

Assuming that work is done by friction, then we have

W(others) = W(f) = [f(k).cosΦ.s] = -f(k).s = -μ(k).m.g.s

Since there's no work done by gravity, U₁ = U₂ = 0

By the time the block comes to rest, K₁ = ½mv² while K₂ = 0

This then mean that

½mv² - μ(k).m.g.s = 0

½v² = μ(k).g.s

v² = 2μ(k).g.s

v = √[2μ(k).g.s]

Solving for v, we have

v = √(2 * 0.2 * 9.8 * 0.31)

v = √1.2152

v = 1.10 m/s

Finally, using law of conservation of momentum equation, we have

V(a) = [(m(a) + m(b) / m(a)] * V

V(a) = [(0.009 + 1.2) / 0.009] * 1.1

V(a) = [1.209 / 0.009] * 1.1

V(a) = 134.333 * 1.1

V(a) = 147.77 m/s

i think the answer is a, hope this is right, that would be my best guess.