A person slaps her leg with her hand, which results in her hand coming to rest i

Question

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 milliseconds from an initial speed of 2.75 m /s. What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.65 kg? Number Would the contact force be any different if the woman clapped her hands together, each with an initial speed of 2.75 m / s, if they come to rest in the same time interval as above O O O O O It depends on the coeffcient of friction between the two hands. No, because the change in momentum for the first hand will be the same. No, because the second hand has zero momentum Yes, because the change in momentum is different Yes, because the initial momentum of the system will be different due to the second hand.

Explanation / Answer

1)Given,

t = 2.65 ms ; u = 2.75 m/s ; m = 1.65 kg

we know from eqn of motion

v = u + at

a = (v - u)/t = (0 - 2.75)/2.65 x 10^-3 = - 1037.74 m/s^2

We know that

F = ma = 1.65 x 1037.74 = 1712.27 N

Hence, F = 1712.27 N

No, because the change in momentum for the first hand will be the same.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.