<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car
change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2)
where v(f) and v(i) are the final and initial speeds
change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2)
185,000=1/2 m(255m^2/s^2)
solving for m
m=1451kg</span>
Where’s the question page at??
I think it’s ultraviolet, if not i’m sorry!
The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
- <em>mass of the bullet, m = 23 g = 0.023 g</em>
- <em>speed of the bullet, u = 230 m/s</em>
- <em>mass of the wood, m = 2 kg</em>
- <em>final speed of the bullet, v = 170 m/s</em>
- <em>coefficient of friction, μ = 0.15</em>
The final velocity of the wood after the bullet hits is calculated as follows;

The acceleration of the wood is calculated as follows;

The distance traveled by the wood after the bullet emerges is calculated as follows;

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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Answer:
Explanation:
Initial kinetic energy of M = 1/2 M vi²
let final velocity be vf
v² = u² + 2a s
vf² = vi² + 2 (F / M) x D
Kinetic energy
= 1/2 Mvf²
= 1/2 M ( vi² + 2 (F / M) x D
1/2 M vi² + FD
Ratio with initial value
1/2 M vi² + FD) / 1/2 M vi²
RK = 1 + FD / 2 M vi²