Question

A bullet of mass 20 grams moving with a velocity of 400 metre per second strikes a tree and goes out from the other side with a velocity of 400 metre per second calculate the work done by the bullet in passing through the tree​

Answer 1

The work done by the bullet in passing through the tree is zero.  

Explanation:

Mass of the bullet, m = 20 g = 0.02 kg

The velocity of the bullet before it strikes the tree, v1 = 400 m/s

The velocity of the bullet after it strikes the and goes out through the tree, v2 = 400 m/s

The formula for work done is given by the change in kinetic energy,

∴ W = K.E.2 – K.E.1 = ½ * m * [v2² – v1²]

Substituting the given values, we get

W = ½ * 0.02 * [400² – 400²] = 0

Thus, the work done by the bullet in passing through the tree is zero.

[Note that: However in real-world scenario the velocity of the bullet will reduce once it crosses the tree due to the work done by frictional force]

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Answer 2

Answer:

The work done by the bullet of mass 20gram passing through the the tree at 400m/s and leaving the tree at 400m/s is 0.

Explanation:

This is a question on the change in kinetic energy.

First we write the formula for getting kinetic energy:

Kinetic energy = ½mv²

Now, we have :

m = mass of the object in kilograms

v = velocity of the object in m/s

In this case we have:

m = 20/1000 = 0.02 Kg

V1 = 400 m/s, V2 = 400m/s

Now, initial kinetic energy = ½ × 0.02 × 400² = 1600 joules

Final kinetic energy = ½ × 0.02 × 400² = 1600 Joules

We now write down the formula for getting work:

Work = final kinetic energy - initial kinetic energy.

Work = 1600 - 1600 = 0 joules

Since initial kinetic energy equals to the final kinetic energy, there's no work done.