when a karate strike hits wooden blocks, the hand undergoes an acceleration of -6500 m/s2. Medical data indicates the mass of the forearm and hand to be about 0.7 kg. What is the force exerted on the hand by the blocks?
To find the force exerted on the hand by the blocks, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a):
F = m * a
In this case, the mass of the forearm and hand is given as 0.7 kg, and the acceleration experienced by the hand is -6500 m/s^2 (negative because it's decelerating). Plugging these values into the formula:
F = 0.7 kg * (-6500 m/s^2)
F = -4550 N
Therefore, the force exerted on the hand by the blocks is -4550 Newtons. The negative sign indicates that the force is acting in the opposite direction of motion, causing deceleration.
To find the force exerted on the hand by the blocks, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given:
Mass (m) = 0.7 kg
Acceleration (a) = -6500 m/s^2 (negative sign indicates acceleration in the opposite direction)
Using the formula:
F = m * a
Substituting the given values:
F = (0.7 kg) * (-6500 m/s^2)
Calculating the force:
F = -4550 N (Newtons)
Therefore, the force exerted on the hand by the blocks is -4550 N (in the opposite direction of acceleration).