Explanation:
Given that,
Initial speed of the man's hand, u = 3.75 m/s
Finally, the hand comes to rest, v = 0
Time interval, t = 2.75 ms
Total mass of the forearm and the hand is, m = 1.75 kg
To find,
The magnitude of force exerted on the leg
Solution,
We know that the force exerted from one object to another is given by using second law of motion as :
F = m × a
a is the acceleration of the object
[tex]F= m\times \dfrac{v-u}{t}[/tex]
[tex]F= 1.75\times \dfrac{0-3.75}{2.75\times 10^{-3}}[/tex]
F = -2386.36 N
Therefore, the force exerted on the leg is 2386.36 N.
