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Work Energy Theorem Question:: A 0.0025 kg bullet traveling straight horizontally at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 0.12 m into the tree’s trunk. What is the initial KE of the bullet? What is the final KE of the bullet? What the the change in KE of the bullet? What is the force exerted?

User OfirD
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Answer:

To solve this problem, we can use the Work-Energy Theorem, which states that the net work done on an object is equal to its change in kinetic energy.

The initial kinetic energy of the bullet can be calculated using the formula:

KE = 0.5 * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.

Substituting the given values, we get:
KE = 0.5 * 0.0025 kg * (350 m/s)^2
KE = 306.25 J

Therefore, the initial kinetic energy of the bullet is 306.25 J.

When the bullet hits the tree, it slows down uniformly to a stop while penetrating a distance of 0.12 m into the tree's trunk. We can assume that the work done by the force of friction between the bullet and the tree is equal to the change in kinetic energy of the bullet.

The final kinetic energy of the bullet is zero because it comes to a stop. Therefore, the change in kinetic energy is:

ΔKE = final KE - initial KE
ΔKE = 0 - 306.25 J
ΔKE = -306.25 J

The negative sign indicates that the kinetic energy of the bullet has decreased.

To calculate the force exerted on the bullet, we can use the formula for work:

W = F * d * cos(θ)
where W is the work done, F is the force, d is the distance, and θ is the angle between the force and the displacement.

Since the force is acting in the opposite direction to the displacement, the angle θ is 180 degrees (cos(180) = -1). Therefore, the formula becomes:

W = -F * d

Substituting the given values, we get:

-306.25 J = -F * 0.12 m
F = 2552.08 N

Therefore, the force exerted on the bullet is 2552.08 N.
User Liran Orevi
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