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A 0.02kg bullet is traveling with a speed of 300m/s when it hits a wall. Assuming it comes to rest after traveling 0.1m with constant deceleration into the wall.
1. Calculate the work done by the wall on the bullet as the bullet comes to a stop.
2. Calculate the deceleration of the bullet as it comes to a stop.

Respuesta :

hamzaahmeds

Answer:

1. W = 900 J

2. a = -450000 m/s²

Explanation:

2.

First, we calculate the deceleration of the bullet by using 3rd equation of motion:

[tex]2as = V_{f}^{2} - V_{i}^{2}\\[/tex]

where,

a = deceleration = ?

s = distance traveled = 0.1 m

Vf = Final Speed = 0 m/s (it eventually stops)

Vi = Initial Speed = 300 m/s

Therefore,

[tex]2a(0.1\ m) = (0\ m/s)^{2} - (300\ m/s)^{2}\\a = \frac{-90000\ m^{2}/s^{2}}{0.2\ m}\\\\[/tex]

a = -450000 m/s² (negative sign shows deceleration)

1.

Now, we calculate the force using Newton's Second Law:

[tex]F = ma\\F = (0.02\ kg)(-450000\ m/s^{2})\\F = - 9000\ N[/tex]

Negative sign shows decelerating force.

For work done:

[tex]Work\ Done = W = Fs\\W = (9000 N)(0.1\ m)[/tex]

W = 900 J

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