Question

A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board. If the time of contact with the board is 0.002 0 s, and the mass of the coordinated hand and arm is 1.0 kg, what is the force exerted on the board

Answer 1

The force exerted on the board by the karate master given the data is -4500 N

Data obtained from the question

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 1 m/s
• Time (t) = 0.002 s
• Mass (m) = 1 Kg
• Force (F) = ?

How to determine the force

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

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