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.

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:

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,

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

1.

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

Negative sign shows decelerating force.

For work done:

W = 900 J

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.