The 4.0-kg head of an ax is moving at 36 m/s when it strikes a log and penetrates 0. 03 m (3 cm) into the log. What is the average force the blade exerts on the log?

Explanation:

To calculate the force we will use newton second law,

F = ma

Where

F is the force

m is the mass of the body

a is the acceleration of the body

Given that the mass is 4kg

Then, applying equation of motion to know the acceleration of the body when it hit the log.

Initially before it hit the log, the velocity of 36m/s

Initial velocity u = 36m/s

Then, finally the object came to rest, then, final velocity is 0m/s

v = 0m/s,

During this period, the object travelled a distance of 0.03m

Then, using equation of motion

v² = u² + 2as

0² = 36² + 2 × a × 0.03

0 = 1296 + 0.06a

-0.06a = 1296

a = 1296 / 0.06

a = -43,200 m/s²

The negative sign shows that the object sea decelerating in the log.

So, a = 43,200 m/s²

Then, applying the Newton second law formula

F = ma

F = 4 × 43,200

F = 172,800 N

The average force exert on the wood is 172,800 N

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